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math/0106167 | It suffices to prove the statement for the normalized complexes. By REF , MATH is a deformation retract of MATH. So, applying the perturbation lemma, we have to show that all the extra terms in the perturbation series vanish. Now the normalized homotopy operator is induced from the original homotopy operator MATH. NAME and NAME show that the operator MATH is universal, in the sense that it is a linear combination (with integral coefficients) of simplicial morphisms of MATH (page REF, NAME REF ). One knows that any order preserving map MATH between finite ordinals can be uniquely decomposed as, MATH such that MATH and MATH and MATH are cofaces and MATH are codegeneracies (compare for example, CITE, page REF). On dualizing this and combining it with the NAME result, it follows that the homotopy operator MATH is a linear combination of operators of the form MATH, where MATH, MATH are vertical and horizontal degeneracy operators and MATH is another operator whose specific form is not important for the sake of this argument. Now we look at the perturbation formula. We are claiming that the induced operator on the normalized chains MATH where MATH is zero. This follows from the above observation since Image MATH . Image MATH degenerate chains. |
math/0106168 | Apply REF of MATH, to obtain : MATH . And REF is obtained by a direct application of the inverse NAME transform. It remains to show that, indeed, the domain MATH is nonempty. However, this fact follows from a special version of NAME 's lemma due to NAME (see for example, CITE), which (adapted to the present context) states that MATH has an admissible solution MATH if and only if MATH is the only solution of the system MATH. In other words, MATH is the only solution of MATH. |
math/0106175 | According to CITE, integer valued multiplicity functions are nonsingular in the sense of CITE. This means (see CITE, p. REF) that there exists a generalized NAME function MATH, a holomorphic MATH-invariant in both variables solution of the system of differential equations MATH . This function is unique up to scaling, and can be normalized by the condition MATH. Now consider the function MATH. This function is a holomorphic invariant solution of the above system, so it must be proportional to MATH. This implies that MATH, as desired. |
math/0106175 | Let MATH be the span of left (or right) tensorands of MATH. This is a graded subspace of MATH. Let MATH be a homogeneous basis of MATH. Then we can write MATH in the form MATH, where MATH is another homogeneous basis of MATH. Thus, for any MATH, we have MATH . But the function MATH is analytic. Thus, MATH cannot have poles and hence is a polynomial. Let us now substitute MATH in the last equality. Since MATH, we get MATH, where MATH. This sum is clearly finite. Thus, MATH, that is, MATH, as desired. |
math/0106175 | REF is clear. Proof of REF : Let MATH be a homogeneous basis of MATH, and MATH the dual basis. Then MATH. Applying MATH to this equation, we get MATH . Substituting MATH, we get MATH so MATH, as desired. Proof of REF : By REF , the right hand side is MATH. Proof of REF : by REF MATH. |
math/0106175 | First of all, MATH clearly contains the NAME vector field, so any submodule of MATH has to be graded. Thus, it is sufficient to show that for any homogeneous element MATH, one has MATH and MATH. But this is clear, since for any homogeneous MATH one has MATH, and MATH for MATH of degree MATH such that MATH (which exists by nondegeneracy of the form). |
math/0106175 | CASE: Looking at the symbols of MATH and using the NAME theorem, we conclude that the dimension cannot be more than MATH (since this is the dimension of the space of ``abstract" solutions of the system, in the sense of differential NAME theory). On the other hand, for generic MATH, it is easy to see that the functions MATH, MATH, are linearly independent elements of MATH. Thus the dimension is generically (and hence always) greater than or equal to MATH. Combining the two results, we get that the dimension is exactly equal to MATH. CASE: The statement says that elements of MATH are MATH-quasiinvariant. This is clear for generic MATH since we showed in the proof of REF that MATH is a basis of MATH. Therefore, it is true for all MATH. CASE: This is clear, as MATH is graded and finite dimensional. |
math/0106175 | This follows from the fact that MATH. |
math/0106175 | It is clear that when MATH is regular then MATH, because in this case MATH form a basis of MATH. Thus, it is sufficient to show that MATH has exactly the prescribed order of vanishing on the reflection hyperplanes. To show this, let MATH be a generic point of MATH, and MATH be a nonzero vector orthogonal to MATH. Define the function MATH . This function is obviously nonzero, and is well defined up to normalization (the normalization depends on MATH). It is easy to see that the functions MATH and MATH, MATH, are linearly independent, and form a basis of MATH. Therefore, the wedge product MATH has a nonzero finite limit as MATH. But it is clear that MATH is a constant multiple of MATH. This implies the required statement. |
math/0106175 | First, note that the pairing MATH is nondegenerate for any MATH. Indeed, since MATH is graded, the degeneracy locus of this pairing in the k-space is invariant under dilations. Also, this locus is clearly closed. So, if it is nonempty, it must contain zero. But the pairing between MATH and MATH is nondegenerate by the definition of MATH. This implies that for any regular point MATH, the evaluation map MATH is an isomorphism (since for MATH, and MATH is a basis of MATH). Thus, MATH is nonzero outside of the reflection hyperplanes in MATH. So it suffices to check that MATH has exactly the predicted degree of vanishing on the hyperplanes, that is, degree MATH on MATH. Let us first check that MATH has degree of vanishing at least MATH on MATH. To this end, look at the limit in which MATH approaches a generic point MATH on a hyperplane MATH. Since MATH are quasiinvariants, for any MATH, the difference between MATH and MATH is of the order at least MATH in this limit. This gives the desired lower bound. Now let us obtain the upper bound. As we mentioned, the pairing MATH given by MATH is nondegenerate. Thus, there exists a basis MATH of MATH such that MATH. Let us express the solutions MATH via this basis. It is clear that MATH, where MATH. Thus, MATH and MATH . The second factor is holomorphic in MATH. Thus, the lower upper bound follows from REF . The Lemma is proved. |
math/0106175 | This is obtained from REF by comparing the degrees of the two sides. |
math/0106175 | Let MATH be regular. Consider the function MATH. It is easy to see from quasiinvariance of MATH that this function (as a function of MATH) extends to a holomorphic function on MATH. In particular, there exists a limit MATH and MATH is an antisymmetric m-harmonic polynomial. Hence, MATH, where MATH. Consider the polyvector MATH (the last expression applies to regular MATH only). We have MATH. Thus, it is sufficient for us to show that MATH is nonzero. For regular MATH, we have MATH where MATH. Thus, by the lemmas on determinants, MATH . But we have seen above that MATH. Hence, MATH, as desired. |
math/0106175 | The first statement is clear from REF . The second statement follows from the first one, since MATH, and MATH. |
math/0106175 | It is sufficient to show that the restriction of MATH to MATH is nondegenerate. But this follows from the fact that MATH is an isomorphism REF , and that MATH is nondegenerate (definition of MATH). |
math/0106178 | Let MATH be the Hermitian matrix defined by MATH. Then MATH, since the zeroth order of MATH is orthonormal with respect to MATH. From CITE we know that there exists a matrix MATH such that MATH. Then MATH is the desired transformation. |
math/0106178 | Let MATH be a good cover of MATH, and let us fix deformed trivialization maps MATH and transition functions MATH as in REF. Let MATH be a deformation quantization of MATH with respect to MATH. Let MATH be a MATH-algebra isomorphism, that, by CITE, can be chosen to preserve supports (see REF ). Such a MATH gives rise to a collection of local maps MATH by MATH, satisfying MATH and MATH (by REF ). It follows from REF that MATH, and therefore MATH . Since MATH is invertible and MATH, we can write (see REF) MATH for some MATH, and MATH. The deformed cocycle REF imply that, on triple intersections MATH, the function MATH must satisfy MATH . This shows that MATH is integral and does not depend on MATH. Since the classical limit of MATH is just the usual addition, we get MATH . But the complex NAME class defined by MATH, viewed as a NAME class, is the NAME class of MATH. Thus MATH. |
math/0106178 | If MATH is a MATH-derivation, then MATH is clearly a MATH-automorphism. For the converse, define MATH. It follows that MATH with some rational coefficients MATH, obtained by recursion. From the fact that MATH is an automorphism, we obtain MATH . This equation can be seen as a fixed point condition for a MATH-linear operator acting on MATH-bilinear maps on MATH, and this operator is clearly contracting in the MATH-adic topology. Thus, by NAME 's fixed point theorem, there exists a unique fixed point, which must be MATH (see for example, CITE). Therefore MATH, and MATH is a derivation. |
math/0106178 | Let MATH be an equivalence, MATH. Then MATH defines a new MATH-involution for MATH of the form MATH, where MATH is a MATH-automorphism. We can write MATH, where MATH is a real derivation of MATH. Thus MATH is still a MATH-automorphism, and the map MATH is a MATH-equivalence between MATH and MATH. |
math/0106178 | Assume that MATH and MATH are NAME equivalent via a line bundle MATH. Equip MATH with a Hermitian fiber metric MATH, and let MATH be a quantization respect to MATH. The endomorphisms MATH form a MATH-algebra strongly NAME equivalent to MATH, see CITE. This algebra is isomorphic to MATH, and, by REF , we can chose the isomorphism to be a MATH-isomorphism. Hence MATH and MATH are strongly NAME equivalent. For the converse, see CITE. |
math/0106178 | The first part is clear. For the second part, let us first consider standard-order. In this case, MATH by REF whence REF is well-defined for MATH. The general case follows from MATH. A local computation shows that REF defines a deformation quantization of MATH. The third part again follows from REF and the fact that MATH. |
math/0106178 | Since REF is obviously satisfied, REF is globally defined. The remaining properties of a deformation quantization of MATH are easily verified from the local formula. Again MATH and REF imply that MATH coincides with MATH on pulled-back sections. Thus the MATH are still orthonormal. |
math/0106178 | Choose local functions MATH such that MATH. Then we know that MATH and REF is a simple computation using REF and the commutativity of all MATH. |
math/0106178 | Let MATH. A straightforward computation shows that MATH, since MATH and MATH is a representation satisfying MATH. Thus the right hand side of REF is a global section. A similar computation shows that MATH, whence MATH is well-defined. From the fact that MATH is a MATH-representation, one obtains for sections/functions with small enough support the relation MATH . Then a partition of unity argument implies that MATH is isometric. Finally we choose MATH and MATH such that MATH on MATH. Then clearly MATH implies surjectivity. This shows the first part. The second part is trivial since MATH is the quotient of MATH by the vectors of length zero. For the third part, we compute locally MATH which is sufficient since all representations are local. |
math/0106180 | Suppose MATH and MATH are two incomparable solutions MATH for frontier conditions MATH. It follows from REF , equalities MATH and MATH, and the inequality MATH that MATH . And by analogy, MATH . Gathering estimates REF in one chain we can see that all inequalities in REF are actually equalities. This fact and REF finish the proof. |
math/0106180 | Let us represent MATH as MATH where MATH. Then suppose that there exists an unordered collection MATH minimizing MATH. For this collection MATH and, therefore, for MATH the inequality MATH holds (remind that all MATH). But, evidently, for ordered Boolean matrices-layers MATH with coordinates MATH we have MATH . Hence, unordered sequence MATH can not minimize MATH. Since for any ordered collection MATH the equality MATH is valid, actually, MATH, and vise versa, each solutions MATH specifies the solution MATH . |
math/0106182 | The proof of this statement is standard and originates in CITE. |
math/0106182 | MATH is a basis of MATH near the origin, where MATH have the form MATH . For any MATH -vector field given by MATH, we compute : MATH . Thus, we have MATH where V is a section of MATH. Now consider a MATH -vector field, that, restricted to MATH, is given by MATH where MATH. MATH . The last equality follows from REF. Using REF , we get MATH . In particular, observe that (MATH below being the unit vector along the ``MATH-axis") MATH since, by complex tangency, MATH. Therefore MATH where MATH . Consider the sesqui-linear forms MATH . Since MATH for each MATH and for each MATH. Thus, MATH are all Hermitian forms that are identically zero. Consequently, MATH for every MATH. Since MATH has maximal rank for each MATH, we conclude that MATH . This implies that, as MATH, MATH. So, MATH . This implies that MATH, whence MATH (since MATH). |
math/0106182 | Without loss of generality, we may assume that MATH is defined by a global defining function MATH that is defined in a neighbourhood MATH. Recall that MATH is of type MATH along MATH and that MATH for each MATH. Pick a MATH. There exist a MATH open in MATH and a real-analytic imbedding MATH. It must be noted that MATH, that is, MATH depends on MATH, but for purposes of notational convenience, we will suppress the dependence on MATH. It can easily be shown, using standard compactness and homogeneity arguments, that by our hypothesis on type along MATH, MATH and for each MATH, there exists a neighbourhood MATH such that REF is true uniformly for all MATH with a uniform constant MATH. By REF , there exist an open subset of MATH, MATH and a smooth family of biholomorphisms, MATH, having the effect that for each MATH, MATH is defined by a MATH as given in REF . Adopting coordinates MATH, write MATH where MATH is the polynomial occuring in REF . Let MATH, consider the ball MATH, and let MATH. Define MATH. Note that MATH. Also MATH. MATH and from REF , we can infer that the above inequality is true uniformly for MATH. Note that the first equality in REF follows from REF . Let MATH be the complexification of MATH (that is, MATH is defined, wherever the resultant power-series converges, by replacing the real variable MATH by the complex variable MATH in the power-series of MATH). Since MATH is a smooth family, choosing MATH appropriately, we can find a MATH such that MATH are all defined as holomorphic maps on MATH for each MATH. Shrinking MATH if necessary, we define MATH by MATH where MATH is the complexification of MATH in an appropriately small neighbourhood of MATH. In what follows, we will write MATH, and MATH. By REF , which says that MATH when MATH, or by the normal form REF in case MATH, MATH has the series expansion MATH . Thus, MATH . Since the above is true uniformly for all MATH with a uniform constant MATH, there is a MATH such that MATH . Another way of saying this is that the complex analytic set MATH meets MATH precisely along MATH. We can, therefore, find an open neighbourhood MATH of MATH in MATH and a complex submanifold MATH of MATH which is the complexification of MATH near MATH. MATH is an open cover of MATH. As MATH is compact, there exist MATH and a tubular neighbourhood MATH of MATH such that CASE: MATH and MATH. CASE: MATH is a complexification of MATH, and a complex submanifold of MATH such that MATH. Let MATH be the real-analytic function prescribed on MATH. Shrinking MATH if necessary, we may assume that MATH extends to a holomorphic function MATH on MATH. MATH has a basis of NAME neighbourhoods. This follows from results by CITE, CITE, and this is where the assumption about MATH being real-analytic gets used. Choose a NAME domain MATH such that MATH is holomorphic on the complex submanifold MATH of MATH. By standard techniques, we can show that MATH extends to a MATH. We remark that this last step reflects a technique used in CITE (which follows from theorems A and B of NAME). Thus, MATH is an analytic interpolation manifold. |
math/0106182 | We assume that MATH is an analytic interpolation manifold. By hypothesis, there exists a real-analytic imbedding MATH onto a simple closed curve MATH, so that the following happens : For MATH sufficiently small, MATH extends to a regular, injective, holomorphic map MATH on MATH. There is a small disc MATH, centered at MATH, such that (defining MATH), without loss of generality, we have MATH . Now choose some MATH in MATH. Define MATH which is real-analytic on MATH, and so extends real-analytically to MATH. This last conclusion follows from a result by CITE. By assumption, there exists a MATH holomorphic in a neighbourhood, call it MATH, of MATH, with MATH. We can choose MATH above to be so small that MATH. We can then define MATH . Clearly MATH and MATH. The latter implies that MATH. Yet, MATH whereas MATH has a pole at MATH. This is a contradiction. Our assumption that MATH is an analytic interpolation manifold must, therefore, be false. |
math/0106183 | Let MATH be a generating set for the collection of entourages in the space MATH, in the sense that every entourage MATH is contained in some finite composite MATH. Without loss of generality, let us assume that each generator MATH contains the diagonal MATH, is symmetric in the sense that MATH, and every finite composite MATH of generators is an entourage. Define MATH to be the set of finite sequences of elements of the set MATH. The set MATH can be ordered; we say MATH if MATH is a subsequence of MATH. Let MATH. Let us say a subset MATH is MATH-sparse if MATH for all points MATH such that MATH. By NAME 's lemma there is a maximal MATH-sparse set MATH. Consider a finite sequence MATH. Form the composite MATH and union MATH, and look at the collection of bounded sets: MATH . Since each of the generators is symmetric and contains the diagonal, the composite MATH contains each of the generators that form it. By maximality of the MATH-sparse set MATH, the collection MATH is a cover of the space MATH. Choose a point MATH. Then the set MATH is a discrete subset of a compact set, and is therefore finite. Thus the cover MATH is good in the sense of REF . We now need to check that the family of covers MATH forms a coarsening family. Consider the family of entourages MATH. By construction, every set MATH takes the form MATH for some point MATH, and every entourage MATH is contained in some entourage of the form MATH. All that remains is to check that for every point MATH and every pair of elements MATH such that MATH there is a set MATH such that MATH. It suffices to look at the special case where we have sequences MATH and MATH in the indexing set MATH. By maximality of the MATH-sparse set MATH we can find a point MATH such that MATH. Hence MATH. |
math/0106183 | Since the map MATH is a coarse map, it takes entourages to entourages. The given conditions on coarsening families ensure that for each element MATH, MATH for some element MATH. Hence we have an induced map of simplicial sets MATH. This map is proper since the map MATH is coarse. By definition of the nerve of a cover, up to proper homotopy the map MATH is independent of any choice of coarsening family. We therefore obtain a homomorphism: MATH by taking locally finite homology groups followed by the direct limit. |
math/0106183 | Without loss of generality suppose we have a coarse homotopy MATH such that MATH and MATH. Let MATH, MATH, and MATH be coarsening families for the spaces MATH, MATH, and MATH respectively. Then for all elements MATH and MATH we can find an element MATH such that MATH. Hence we have an induced proper map of simplicial sets: MATH . Now, let us write the cover MATH of the generalised ray MATH as a sequence of bounded sets MATH where MATH and MATH. Then we can define a continuous map MATH by sending a natural number MATH to the vertex MATH and a point MATH to the appropriate point on the edge joining the vertices MATH and MATH. Hence we have a proper continuous map MATH such that for each point MATH the function MATH is eventually constant. The induced maps MATH and MATH are thus properly homotopic; we obtain the appropriate proper homotopy by renormalising the map MATH. The maps MATH and MATH therefore induce the same map at the level of locally finite homology. Taking direct limits, the maps MATH and MATH must be equal. |
math/0106183 | Let MATH be a coarsening family for the space MATH. Then the spaces MATH and MATH have coarsening families MATH and MATH defined by writing: MATH respectively. Write MATH, MATH, and MATH. The decomposition MATH is coarsely excisive, so if we look at interiors then: MATH . By the existence of NAME sequences in ordinary homology (see for example CITE) we have natural maps MATH and exact sequences: MATH . Taking direct limits, we have an exact sequence: MATH as required. |
math/0106183 | Let MATH and MATH be partitions of unity subordinate to the open cover MATH. Then the resulting maps MATH and MATH are properly homotopic, and so induce the same maps at the level of locally finite homology. |
math/0106183 | We have a commutative diagram of exact sequences: MATH . The excision map MATH is an isomorphism. Write MATH. Then a diagram chase tells us that we have a long exact sequence: MATH as desired. |
math/0106183 | If MATH is a continuous map of pairs there is an induced continuous map MATH. By REF we can define a coarse map MATH for some radial contraction MATH. The coarse homotopy type of the composition MATH does not depend on the choice of contraction MATH so we obtain a functorially induced map MATH. Let MATH be a relative homotopy. Choose a radial contraction MATH such that each map MATH is a coarse map. Then we can define a coarse homotopy MATH by the formula: MATH for all points MATH, MATH, and MATH. Hence the maps MATH and MATH are relatively coarsely homotopic, so the induced maps MATH and MATH are equal. If MATH is a pair of subspaces of the sphere MATH then MATH is a pair of coarse spaces so we have natural maps MATH and a long exact sequence MATH where the maps MATH and MATH are induced by the inclusions MATH and MATH respectively. Finally, suppose we have a pair MATH and a subset MATH such that MATH. Since the space MATH is compact we can find a real number MATH such that the neighbourhood MATH is a subset of the space MATH. Hence for all MATH there exists MATH such that: MATH . Looking at cones, we see that the decomposition MATH is coarsely excisive, and the excision axiom follows. This completes the proof. |
math/0106183 | The topological space MATH can be compactified by adding a `hemisphere at infinity'. The coarse space MATH can be viewed as the space MATH with coarse structure defined by looking at continuous control with respect to this compactification. As in REF we can define a coarse map MATH by writing MATH. The composition MATH is equal to the identity MATH, and we can define a coarse homotopy linking the functions MATH and MATH by the formula: MATH . |
math/0106183 | The maps MATH and MATH are injective, and the coarse structures of the images MATH and MATH are defined to be those inherited from the spaces MATH and MATH, respectively, under these maps. This establishes the first part of the proposition. Certainly the space MATH is equal to the union MATH. Let MATH be an entourage for the space MATH. We want to find an entourage MATH such that: MATH . By definition of the coarse structure on the space MATH we can write: MATH where MATH and MATH are entourages for the spaces MATH and MATH respectively, MATH and MATH are bounded subsets, and MATH is the canonical quotient map. Observe that: MATH . Let MATH and MATH be the diagonals in the spaces MATH and MATH respectively. Consider any point MATH and let MATH be an entourage containing the image: MATH . Then: MATH and we are done. |
math/0106183 | Let MATH denote the space MATH equipped with the coarse structure inherited from the generalised ray MATH and let MATH be the space MATH. Then the space MATH is coarsely equivalent to the space MATH, and therefore coarsely homotopy-equivalent to the generalised ray MATH. Let MATH. Then the space MATH is coarsely equivalent to the space MATH. The space MATH can be written as the union MATH, and the intersection MATH is coarsely equivalent to the coarse sphere MATH. Hence by REF we have a NAME sequence MATH as required. |
math/0106183 | Let us write a given coarse MATH-sphere, MATH, as a coarsely excisive union, MATH, of two generalised rays. The intersection, MATH, is bounded and is therefore coarsely equivalent to a single point, MATH. Considering NAME sequences we have a commutative diagram: MATH . The rows in the above diagram are exact. With the possible exception of the map MATH, the vertical arrows are isomorphisms. Hence the map MATH is also an isomorphism by the five lemma. Now suppose that the map MATH is an isomorphism for every coarse MATH-sphere. Let MATH be a coarse MATH-sphere. Then we can write MATH as a coarsely excisive union, MATH, of two MATH-cells, with intersection coarsely equivalent to some coarse MATH-sphere, MATH. The result now follows by induction if we look at NAME sequences and apply the five lemma as above. |
math/0106183 | The map MATH is certainly an isomorphism if the space MATH is a coarse MATH-complex with just one cell. Suppose that the map is an isomorphism whenever the space MATH is a coarse MATH-complex with fewer than MATH cells. Let MATH be a coarse MATH-complex of dimension MATH that has MATH cells. Let MATH be a cell of dimension MATH and let MATH be a coarse MATH-complex such that the MATH-complex MATH is obtained from MATH by attaching the cell MATH. By REF we have a commutative diagram: MATH where MATH is a coarse MATH-sphere, MATH is a generalised ray coarsely homotopy-equivalent to the cell MATH, and the rows are NAME sequences. All of the vertical arrows except for possibly the map MATH are isomorphisms by inductive hypothesis and REF . Hence by the five lemma the map MATH is also an isomorphism. Thus the result holds for any finite coarse MATH-complex by induction. By coarse homotopy-invariance the result follows for any space coarsely homotopy-equivalent to a finite MATH-complex. |
math/0106183 | By REF the map MATH is an isomorphism whenever the space MATH is a finite coarse MATH-complex. The result now follows by looking at long exact sequences and applying the five lemma. |
math/0106183 | Let MATH be an open coarsening family on the space MATH. Recall from the comments at the end of REF that we can define a proper continuous map MATH by the formula: MATH . The coarsening map MATH is defined to be the direct limit of the induced maps MATH. The map MATH can be defined to be the direct limit of the maps: MATH . |
math/0106183 | The MATH-algebra MATH consists of all operators on some infinite-dimensional NAME MATH-module. An NAME swindle argument tells us that the MATH-theory groups of such a MATH-algebra are all trivial. |
math/0106183 | The generalised ray MATH is flasque in the sense described in REF. Hence, by REF, the MATH-theory groups MATH are all trivial. |
math/0106183 | We can find a coarsening family MATH for the space MATH such that each nerve MATH is properly homotopic to the ray MATH. An NAME swindle argument tells us that the groups MATH are all trivial. Taking direct limits, the groups MATH must also be trivial. |
math/0106183 | The result is immediate from the previous three lemmas and REF . |
math/0106185 | These results are implicit in the work of NAME - CITE (and slightly more explicit in CITE). The existence of the filtration is exactly CITE; that we can order the bipartitions MATH by dominance follows from the choice of MATH as a maximal chain of ideals in the proof of this result. REF follows by combining REF. |
math/0106185 | Let MATH be an embedding into a complete discrete valuation ring MATH where MATH has residue field MATH and MATH is a field extension such that MATH. Since MATH is absolutely irreducible, MATH. Hence it is enough to prove the statement under the assumption that MATH is a complete discrete valuation ring. Set MATH and let MATH where MATH is the trivial MATH - module. Then MATH is a projective MATH - module by the definition of the MATH. Therefore, by REF applied to MATH, the module MATH has a NAME filtration. Let MATH; then, by assumption, MATH; therefore, MATH, where MATH. Consequently, there is a surjective homomorphism MATH; tensoring with MATH gives a surjective homomorphism MATH (as MATH is the head of MATH). Since MATH is a projective MATH - module there exists a map MATH which makes the following diagram commute. MATH . Now, MATH is surjective so MATH must also be surjective. On the other hand, MATH is a projective MATH - module so we also have a commutative diagram MATH . Hence, MATH is a direct summand of MATH. Note that MATH thus, MATH. However, MATH; so we have shown that the modular reduction of MATH is MATH, as required. |
math/0106185 | By definition, MATH and MATH are in the same block if and only if there exists a sequence of bipartitions MATH such that MATH and MATH have a common composition factor. Thus, it is enough to prove that if MATH appears in MATH then MATH and MATH are linked by hooks. If MATH then MATH by REF . The sum formula implies that MATH appears in MATH for some MATH such that MATH and MATH and MATH are linked by hooks. By induction on dominance MATH and MATH are linked by hooks so we are done. |
math/0106185 | By REF we may assume that MATH. Further, by REF it is enough to show that one block of MATH has infinite representation type; we will show that the block MATH with residues MATH has infinite representation type. There will be several cases to consider. To begin suppose that MATH. Because all of the residues in MATH are distinct a bipartition MATH appears in MATH only if MATH and MATH are both hook partitions; that is, MATH for some MATH. Define bipartitions MATH . It is easily checked that all of these bipartitions belong to MATH. Certainly, the two sets of bipartitions MATH and MATH are disjoint; the restrictions on MATH and MATH ensure that MATH and MATH for any MATH. Consequently, this is a complete list of the bipartitions which appear in MATH, with no repeats. Suppose that MATH and that MATH. CASE: The complete set of NAME bipartitions in MATH is MATH . CASE: For MATH we have MATH . CASE: For MATH and MATH we have MATH . It is easy to see that the bipartitions MATH and MATH, for MATH, are not NAME. Next suppose that MATH. Then a straightforward computation shows that MATH . Note that MATH has two addable MATH - nodes, an addable MATH - node and an addable MATH - node which is a MATH - node if MATH, and when we add an addable MATH - node, there is no removable MATH - node. This shows that MATH is NAME for MATH; the formula for MATH now follows from REF (and the remarks after REF ). The remaining cases are similar: for MATH with MATH we have MATH and we compute MATH for MATH with MATH and MATH compute MATH for MATH with MATH and MATH compute MATH and, finally, for MATH with MATH compute MATH . In each case, an application of REF now completes the proof. By REF in order to prove REF it is enough to show that MATH is not isomorphic to MATH for any MATH. We need to consider several cases. First we observe that MATH by REF . We will use this to determine the structure of MATH. REFMATH and MATH. Then MATH since MATH. By REF we can use REF to compute the decomposition numbers MATH for the four NAME modules appearing in REF ; this gives the following table (omitted entries are zero). MATH . Consequently, MATH. By REF , MATH has submodule isomorphic to MATH. Therefore, MATH has both a simple head and simple socle; so, looking at the submatrix of the decomposition matrix above, the NAME structure of MATH is MATH . We also have MATH . Considering the dual of MATH and MATH, we conclude that MATH and MATH appear in MATH. If MATH is an exact sequence then, for MATH, MATH acts invertibly on MATH so that MATH acts as MATH on MATH. Similarly, MATH also acts as multiplication by a scalar on MATH since MATH. Therefore, every such exact sequence splits and so MATH; consequently, MATH is not a composition factor of MATH. Again, by REF MATH has a NAME filtration so MATH has a NAME filtration whose successive quotients are MATH and MATH. This means that MATH. We now use the fact that MATH to prove that MATH is contained in MATH. As MATH has a unique head there is a surjection MATH. Note also that MATH has a NAME filtration with successive quotients MATH and MATH, MATH, MATH. Since MATH, MATH, MATH must map to MATH, and each of these has unique head which is not isomorphic to MATH, this surjection induces a map MATH. Therefore, MATH contains MATH as a summand. Let MATH be a module such that MATH and MATH and set MATH . Then there is a short exact sequence MATH and MATH. Hence, we also have MATH and consequently MATH. Therefore, MATH contains MATH and it follows that MATH contains MATH as a MATH - submodule, that another MATH appears as the head of MATH, and that these are the only ways in which MATH appear as a composition factor of MATH . Therefore, we have that MATH. Hence, MATH is not of finite representation type by REF , so MATH has infinite representation type by REF MATH and MATH. By REF , we have the same table as above if we replace MATH by MATH. Thus MATH . The argument used in REF shows that MATH contains MATH as a MATH - submodule. Hence, MATH is again of infinite type. REFMATH. Again using REF we find that MATH where MATH if MATH and MATH otherwise. Almost the same argument as before again shows that MATH contains MATH as a MATH - submodule; however, to show that MATH contains MATH we need to argue in a different way: since MATH and MATH, MATH do not contain MATH, the surjection MATH induces a surjection MATH. Thus, once again MATH is of infinite representation type by REF MATH. It remains to consider the case MATH. This case is somewhat degenerate as MATH is the only possible eigenvalue for the action of MATH upon a representation. Because MATH there are no bipartitions of the form MATH in MATH so REF cannot hold in this case. A similar argument shows that we have the following simpler statement when MATH. Suppose that MATH. CASE: MATH is the complete set of NAME bipartitions in MATH. CASE: For MATH we have MATH . To apply this result we do not have to work as hard as in the previous cases. First observe that for MATH the module MATH can be constructed by letting MATH act as MATH on the simple MATH - module MATH (by induction both modules have the same dimension; namely, MATH). Let MATH be the MATH - module obtained by letting MATH act as MATH on the MATH - module MATH. Then MATH cannot be semisimple as a MATH - module because MATH does not act as a scalar. On the other hand, the socle of MATH is simple (being isomorphic to MATH); hence, MATH is indecomposable. This implies that MATH. Now MATH for MATH by REF . Hence, MATH is indecomposable and MATH for MATH. As MATH is self - dual, by taking duals we also have MATH for MATH. Combining these facts we conclude that MATH and that MATH. Therefore, the separation diagram of MATH contains MATH . Consequently, MATH has infinite representation type by REF . This completes the proof of REF . |
math/0106185 | By REF it is enough to show that MATH has infinite representation type when MATH. Also, by the last section we may assume that MATH. Let MATH be the block MATH with residues MATH; so MATH contains the bipartition MATH. We will show that MATH has infinite representation type. To do this we need to consider two cases separately. REFSuppose that MATH and MATH. We consider the block MATH with residues MATH. It is easy to see that there are precisely six bipartitions in this block; namely, MATH, MATH, MATH, MATH, MATH and MATH. Furthermore, of these bipartitions only MATH and MATH are NAME. Since MATH by REF the transpose of the decomposition matrix of MATH is MATH . Let MATH and MATH. We first observe that MATH is the induced module MATH; this follows from a calculation in the NAME group. Similarly, we also have that MATH. Further, MATH is uniserial of length MATH and MATH, for MATH. Next note that MATH, for MATH; hence, the branching rules for NAME modules imply that MATH and MATH. Next note that there are surjective homomorphisms MATH for MATH. As MATH is self - dual, MATH is a uniserial module whose top and bottom is isomorphic to MATH. To conclude, each MATH has a submodule MATH such that both MATH and MATH are isomorphic to the uniserial module MATH. Now we prove that MATH. As MATH, we know that the radical series of MATH has the form MATH . Assume that MATH. Then MATH must have the form MATH . Consider a homomorphism MATH. The assumption that MATH together with the radical structure of MATH imply that this map lifts to a homomorphism MATH, whose image is of the form MATH . This contradicts our assumption that MATH. Hence, MATH and MATH are both non - zero. Therefore, MATH and MATH both contain MATH as a MATH - submodule. Therefore, the separation diagram of MATH is MATH. Consequently, MATH is of infinite type, by NAME 's theorem, and MATH has infinite representation type by REF Suppose that MATH. Now we are considering the block MATH of MATH with residues MATH. Notice that MATH is a residue in MATH only if MATH and MATH is a residue in MATH only if MATH; however, MATH because MATH and MATH; so MATH and MATH cannot be residues in MATH. Consequently, all of the bipartitions in MATH have the form MATH. Define bipartitions MATH, MATH, MATH, MATH and MATH. We will show that MATH has infinite representation type by computing MATH. The bipartitions MATH and MATH are all NAME. Furthermore, MATH and the first five rows of the decomposition matrix of MATH are as follows (all omitted entries are zero): MATH . We use REF to compute MATH, for MATH. We find that MATH . As in REF , by REF this shows that MATH is NAME, proves the formula for MATH and thus gives the second column of the decomposition matrix of MATH. The remaining claims follow from the following calculations, which we leave to the reader. MATH . Here, MATH indicates a linear combination of more dominant terms and if MATH then we replace MATH by MATH. Consequently, MATH. Now, by REF , MATH has a submodule isomorphic to MATH. The composition factors MATH and MATH of MATH cannot appear in MATH because then MATH would appear in the head of MATH. On the other hand, because MATH and MATH are indecomposable, MATH appears in MATH. Recall, again, that MATH has a NAME filtration with successive quotients MATH and MATH. Therefore, the MATH - modules MATH and MATH which appear in MATH are composition factors of MATH and MATH. Since MATH and other MATH are the head and the socle of MATH, MATH does not appear in MATH. Thus MATH and we see that MATH contains MATH as a MATH - submodule by the same argument as before. Consequently, MATH and MATH is of infinite representation type by REF . Therefore, MATH has infinite representation type by REF . This completes the proof of REF . |
math/0106185 | REF both follow from the fact mentioned earlier that the number of ones in a region to the left of a bar is equal to the number of zeros in the region to the right of the same bar. REF is true by definition. Finally, REF follows by counting the rim hooks in MATH (recall that the rim hooks in the path sequence for a partition correspond to subsequences of the form MATH); this is the only place where we use the restriction on MATH. |
math/0106185 | Fix a block MATH of MATH and recall from REF that two simple modules belong to the same block only if they have a common multiset of residues. Note that because MATH the number of distinct residues contained in the diagram of MATH is strictly less than MATH; consequently, we can find a MATH, with MATH, such that MATH is not a residue in MATH. Suppose that MATH appears in MATH. Then the contents of all of the nodes in MATH are contained in the interval MATH. REFSuppose that MATH. As MATH is not a residue for the block MATH the contents of the nodes in MATH are all contained in the interval MATH - note that MATH. Therefore, the multiset of residues for the block MATH is the same as the multiset of contents for MATH. Consequently, we can unwrap a rim hook from MATH and wrap it back on again without changing the residue of the foot node only if the resulting bipartition MATH has the same multiset of contents as MATH. Recall that unwrapping a rim hook from a partition is the same as swapping the ends of a MATH subsequence to give MATH and that wrapping a hook back on changes some MATH into MATH. Now, the contents of a bipartition MATH are determined by the partial sums in the path sequence of MATH; because of this, the only way to unwrap a rim hook from MATH and then wrap it back on to give a bipartition MATH with the same multiset of contents is by interchanging some MATH and MATH in the path sequence: MATH . Moreover, MATH if and only if MATH moves to the left. (Note that MATH in this case as the number of MATH subsequences in the two path sequences is the same.) If MATH then the path sequence for MATH does not contain any MATH's so by the sum formula, REF , and by REF , the NAME module MATH is the only simple module in the block MATH. In particular, MATH is semisimple and so of finite type in this case. If MATH then MATH by REF . In this case by REF the block MATH contains at most MATH bipartitions; namely, the bipartitions MATH whose path sequences contain exactly one MATH and MATH's and which agree with the path sequence for MATH on all of the MATH's and MATH's. By ordering these bipartitions according to the location of the (unique) MATH in their path sequence we may assume that MATH (for example, MATH occupies the leftmost position in the path sequence of MATH and the rightmost position in MATH). For our purposes, it is enough to prove that MATH and MATH for MATH; in particular, we do not need to know that MATH, for MATH. In fact these modules are always non - zero; we include the proof below because it yields the remarkable fact that when MATH the number of NAME modules belonging to a block is either MATH, MATH or MATH. The removable nodes in a partition correspond to the MATH subsequences in the path sequence. Suppose that MATH. Then MATH contains a removable node MATH; furthermore, this node is automatically good because if MATH is a MATH - node then there is no addable MATH - node below MATH because MATH. Let MATH be the bipartition with MATH. Then MATH is NAME because either the path sequence for MATH contains a MATH (so MATH is NAME by induction on MATH), or MATH (by the second last paragraph); hence, MATH is NAME. On the other hand, MATH is not NAME because either we can apply induction after removing a node from MATH, or the path sequence for MATH is MATH in which case it is easy to see that MATH is not NAME. Now MATH by REF since MATH for MATH. To complete the proof we claim that MATH, for MATH, and MATH. To see this we apply the sum formula. As discussed earlier, the leg length of a hook MATH in a path sequence is given by the number of MATH's strictly contained in the subsequence. Consequently, when we unwrap the hook MATH from MATH and wrap it back on to give some MATH then, modulo MATH, the difference in the leg lengths of the two rim hooks is equal to the number of MATH's which are strictly contained in the subsequence for the rim hook. Therefore, by REF , for MATH . As we already know that MATH are NAME, and that MATH is not, our claim now follows by induction on MATH. Consequently, the decomposition matrix of the block MATH is MATH and MATH has finite representation type by REF Suppose that MATH. In this case the contents of the nodes in MATH are contained in the interval MATH. Renormalizing MATH as MATH the relation for MATH becomes MATH. The NAME module MATH is relabelled as MATH and the residues in MATH are all changed by adding MATH. Consequently, the residues for MATH are all contained in the interval MATH. Therefore, with this renormalization, the multiset of residues for MATH is the same as the multiset of contents for MATH. Consequently, we can repeat the argument of REF to deduce that decomposition matrix for MATH has the form above; so, again, MATH has finite representation type by REF . |
math/0106186 | It is well-known that a NAME domain MATH can be built up using REF-, REF- and REF-handles only; for such manifolds the surjectivity of MATH is obvious. This surjection now trivially implies that MATH is also a surjection, hence MATH. |
math/0106186 | It is a standard fact that the KREF-surface MATH contains three disjoint copies of the nucleus MATH (recall that MATH); and the intersection form of the manifold MATH is negative definite and nonstandard. So if MATH is negative definite with MATH then REF-manifold MATH we get by replacing the nuclei in MATH by MATH is a smooth manifold with nonstandard negative definite intersection form. The existence of such a manifold, however, contradicts NAME 's REF , showing that MATH is not negative definite. |
math/0106186 | Let MATH be a NAME filling of MATH and consider the NAME embedding MATH where MATH is a minimal surface of general type - the existence of such an embedding is guaranteed by REF . Recall that we can assume that MATH is nonspin with MATH. The product formula of REF shows that MATH: We use the fact that MATH, as a minimal surface of general type has only two basic classes MATH; therefore MATH is nondivisible, but since MATH is nonspin, REF is not characteristic and so MATH. Notice that this computation shows that MATH is the unique basic class for MATH, in particular, MATH is spin. (The product formula of REF applies since MATH by REF and MATH by REF .) Now consider MATH - as before, MATH stands for the KREF-surface. The product formula shows that MATH, and easy handle calculus verifies that MATH is simply connected: MATH is simply connected and we can build MATH on the top of it by adding only REF-, REF- and REF-handles, since MATH is NAME. In conclusion, for the simply connected spin manifold MATH we have that MATH; applying REF of NAME and NAME this fact implies that MATH and since MATH, we get that MATH. |
math/0106186 | Fix a contactomorphism between MATH and the boundary of the NAME domain of REF . For determining MATH consider MATH attached along MATH with framing MATH. The result is a NAME filling of MATH, and according to REF it has NAME characteristic REF. (Since we glue REF-handles along Legendrian curves with framing MATH, the existence of a NAME structure on MATH follows from by now standard arguments discussed in CITE.) Removing the three REF-handles from MATH we arrive to the conclusion MATH. This fact implies that MATH, since MATH and MATH would imply MATH. Therefore MATH admits unramified covers of any degree. Notice that since MATH by REF (and so MATH), we get that MATH. Consider a REF-fold unramified cover of MATH - the result is a NAME filling MATH of REF-fold cover of MATH, which is MATH again. Therefore all the above said - in particular MATH - holds for MATH. Since MATH, we conclude that MATH. Finally we show that MATH for a NAME filling MATH of MATH. According to REF the map MATH is onto - we claim that the (image of the) circles MATH and MATH of REF remain essential in MATH while MATH becomes REF. Since MATH can be given as a pull-back of MATH under unramified cover along the MATH coordinate, if MATH in MATH then a corresponding MATH-fold cover of MATH provides a NAME filling of MATH equipped with MATH. This contradicts REF once MATH, hence MATH in MATH. If MATH in MATH then attaching a REF-handle along MATH with MATH we get a NAME domain MATH with MATH of REF and MATH, since the surface in MATH with boundary MATH together with the core of the handle and the dual torus of MATH in MATH give a hyperbolic pair in MATH. This fact contradicts REF , therefore MATH in MATH. The role of MATH is analogous, hence MATH in MATH. It follows now that MATH admits a CW decomposition with two REF-cells, and so the number of REF is one in this decomposition (since MATH). Since MATH is Abelian (being a factor of MATH), we get that the attaching circle of this REF-cell is homologically trivial, therefore MATH. In particular, MATH. This implies MATH, and so the intersection form MATH can be easily identified with MATH. The proof is now complete. |
math/0106186 | Since MATH admits a unique tight contact structure, adding a REF-handle and a REF-framed REF-handle (as shown by REF ) to MATH results a NAME filling MATH of MATH. Since MATH, we obviously have MATH. This again shows that MATH, which implies MATH, since - according to REF - the fundamental group MATH is the factor of MATH. Now MATH trivially follows. |
math/0106186 | Consider the NAME embedding MATH where MATH is a minimal surface of general type with MATH. Since a REF-manifold with positive scalar curvature metric cannot divide a REF-manifold with nonzero NAME invariants into two pieces both with MATH, the lemma follows. |
math/0106186 | The fact that MATH follows from REF . Recall that the NAME equations admit a unique (up to gauge equivalence) solution on MATH. Now consider an embedding MATH where MATH is a minimal surface of general type. Grafting solutions for the spin-MATH structures MATH and MATH together we get REF basic classes unless MATH or MATH. Since MATH has exactly two basic classes and MATH is nonspin (therefore MATH) we get that MATH, consequently MATH is spin. |
math/0106186 | It can be easily verified that the negative definite MATH-plumbing MATH (see REF ) embeds in the KREF-surface MATH. For a NAME filling MATH consider MATH. Since MATH is spin (and MATH admits a unique spin structure) it is spin and MATH. Since MATH is negative definite, we have that MATH. NAME 's Theorem (see REF ) shows that MATH; since NAME 's famous REF excludes MATH, we conclude that MATH, proving REF for MATH. The part of REF about MATH follows from the fact that MATH does not bound negative definite REF-manifold at all: If MATH and MATH then MATH violates NAME 's REF . |
math/0106190 | By compactness of MATH, the metric spaces MATH have a uniform injectivity radius - that is, there exists MATH such that for each MATH every homotopically nontrivial closed curve in MATH has length MATH, and it follows that every closed MATH ball in MATH lifts isometrically to MATH. Let MATH be the set of pairs MATH such that for some MATH we have MATH. Evidently MATH acts cocompactly on MATH, and so we have a finite supremum MATH . Given MATH and MATH, choose a MATH - geodesic MATH from MATH to MATH and let MATH be a monotonic sequence along MATH such that MATH for MATH and MATH. For any MATH it follows that: MATH . Setting MATH and MATH the lemma follows. |
math/0106190 | For the proof we review briefly notions of extremal length, in the classical setting of simple closed curves, as well as NAME 's extension to the setting of measured foliations CITE. Recall that for any conformal structure on an open annulus MATH there is a unique Euclidean annulus of the form MATH conformally equivalent to MATH, with MATH; the modulus of MATH, denoted MATH, is defined to be the number MATH. For any NAME surface MATH and any isotopy class of simple closed curves MATH, the extremal length MATH is the infimum of MATH taken over all annuli MATH whose core is in the isotopy class MATH. NAME proved CITE that the function MATH defined by MATH extends continuously to a function MATH. Moreover, for any transverse pair of measured foliations MATH with associated conformal structure MATH and quadratic differential MATH, we have MATH . Given MATH, the extremal length horoball based at MATH is defined to be MATH. Note for example that, setting MATH, for every MATH the extremal length of MATH at points of MATH decreases strictly monotonically to zero as the point moves towards MATH, and so every NAME geodesic with positive direction MATH eventually enters MATH in the positive direction and, once in, never leaves. Given MATH, there is a one parameter family of extremal length horoballs based at MATH, namely MATH for all MATH such that MATH. For the first sentence of the theorem, consider two geodesic rays MATH, MATH such that MATH. Pick any extremal length horoball MATH based at MATH. The proof of REF shows that MATH is bounded. However, MATH is an infinite subray of MATH, and moreover as a point MATH travels to infinity in MATH the horoball MATH contains a larger and larger ball in MATH centered on MATH. It follows that MATH and MATH have infinite NAME distance in MATH. The second sentence follows from the first, by dividing each line into two rays. |
math/0106190 | REF is an easy consequence of REF , as follows. Choose a compact subset MATH whose image in MATH covers MATH and such that over any point of MATH there exists a point MATH such that MATH. It follows that the points MATH in REF may be translated to lie in MATH. Identifying MATH diffeomorphically with MATH, compactness of MATH produces a compact family of hyperbolic metrics on MATH, and compactness of the restriction of MATH to MATH produces a compact family of singular Euclidean metrics. Now apply REF . For REF , note that by compactness of MATH and of the compactness of the restriction of MATH to MATH, there exists a uniform MATH such that any hyperbolic metric and any normalized singular Euclidean structure determined by an element MATH has a MATH - hyperbolic universal cover. REF is now a direct consequence of REF . |
math/0106190 | Any infinite order element of a word hyperbolic group has source - sink dynamics on its NAME boundary, and so MATH has source - sink dynamics on MATH. It follows that MATH has an axis in MATH. But the elements of MATH having an axis in MATH are precisely the pseudo-Anosovs CITE. |
math/0106190 | The NAME boundary of a word hyperbolic group is the closure of the fixed points of infinite order elements in the group, and so by REF the set MATH is the closure of the fixed points of the pseudo-Anosov elements of MATH. REF now follows from REF. To prove REF , let MATH . REF says that MATH acts properly discontinuously on MATH, and so it suffices to prove that MATH. Each point MATH is the ideal endpoint of a cobounded geodesic ray, which implies that MATH is uniquely ergodic and fills the surface CITE, and so if MATH then MATH. |
math/0106190 | Let MATH be the limit set of MATH, with weak hull MATH, and note that we trivially have MATH. Note that MATH acts properly on MATH. Indeed, MATH acts properly on MATH, and so any subgroup of MATH acts properly on any subset of MATH which is invariant under that subgroup. Since MATH, and since MATH acts cocompactly on MATH, it follows that MATH is contained with finite index in MATH. This implies that MATH. To complete the proof we only have to prove the reverse inclusion MATH. Given MATH, suppose that MATH, and choose finite index subgroups MATH such that MATH. By the definition of convex cocompactness it follows that MATH. Since MATH it follows that MATH. |
math/0106190 | The key ingredients in the proof are results of NAME from CITE concerning projections from balls and horoballs in MATH to geodesics in MATH, and results of NAME - CITE characterizing MATH - hyperbolicity of proper geodesic metric spaces in terms of projections properties to paths. To begin with, note that the implications MATH are obvious. We now prove that MATH. Suppose we have an orbit MATH of MATH and a constant MATH, and for each MATH we have two points MATH, endpoints of a unique geodesic segment MATH in MATH, such that MATH, MATH, and MATH. The set MATH maps to a single point in MATH and so the projection of MATH to MATH is a bounded set MATH. It follows that each MATH is MATH - cobounded. Now consider an arbitrary orbit MATH of MATH; we must prove that MATH is quasiconvex in MATH. The orbits MATH have finite NAME distance MATH in MATH. Given MATH, choose MATH within distance MATH of MATH, respectively, and consider the geodesic segment MATH and the piecewise geodesic path MATH . Of the five subsegments of MATH, all but the middle subsegment have length MATH, and it follows that MATH is a MATH - quasigeodesic in MATH, with MATH depending only on MATH. Since the geodesic MATH is MATH - cobounded we can apply the following result of CITE to obtain MATH, depending only on MATH and MATH, such that MATH. For any bounded subset MATH of MATH and any MATH there exists MATH such that if MATH is a MATH quasigeodesic in MATH with endpoints MATH, and if MATH is MATH - cobounded, then MATH. It follows that MATH, proving quasiconvexity of MATH in MATH. CASE: Fix an orbit MATH of MATH, and so MATH is quasiconvex in MATH. Let MATH be the set of all geodesic segments, rays, and lines that are obtained as pointwise limits of sequences of geodesics with endpoints in MATH. Let MATH be the union of the elements of MATH. The left action of MATH on MATH is evidently cobounded. By quasiconvexity of MATH it follows that the action of MATH on the union of geodesic segments with endpoints in MATH is cobounded, which implies in turn that the action of MATH on MATH is cobounded. Since MATH is closed and MATH is locally compact, it follows that the MATH action on MATH is cocompact. The set MATH therefore projects to a compact subset of MATH which we denote MATH. All geodesics in MATH are therefore MATH - cobounded. Let MATH be equipped with the restriction of the NAME metric. Note that while MATH is not a geodesic metric space, it is a quasigeodesic metric space: there exists MATH such that any MATH are within distance MATH of points MATH, and the geodesic MATH is contained in MATH. To prepare for the proof that MATH is word hyperbolic, fix a finite generating set for MATH with NAME graph MATH, and fix a MATH - equivariant map MATH taking the vertices of MATH to MATH and taking each edge of MATH to an element of MATH. Since MATH acts properly and coboundedly on both MATH and MATH, and since both are quasigeodesic metric spaces, it follows that the equivariant map MATH is a quasi-isometry between MATH and MATH; pick a coarse inverse MATH. By definition the group MATH is word hyperbolic if and only if the NAME graph MATH is MATH - hyperbolic for some MATH. Our proof that MATH is word hyperbolic will use a result of NAME and NAME, REF: Let MATH be a geodesic metric space and suppose that there is a set of paths MATH in MATH with the following properties: CASE: There exists MATH such that for any MATH with MATH there is a path in MATH joining MATH and MATH. CASE: There exist MATH, and for each path MATH in MATH there exists a map MATH such that: CASE: For each MATH we have MATH. CASE: If MATH then MATH. CASE: If MATH and MATH then MATH . Then MATH is MATH - hyperbolic for some MATH. To prove that MATH is MATH - hyperbolic we take MATH to be the set of geodesic segments in MATH, and we look at the set of paths MATH in MATH. Using some results of CITE, we will show that MATH satisfies the hypotheses of REF . Then we shall pull the hypotheses back to MATH and apply REF . The first result of NAME that we need is the main theorem of CITE: For every bounded subset MATH of MATH there exists MATH such that if MATH is any MATH - cobounded geodesic in MATH then the closest point projection MATH satisfies the MATH contracting projection property with MATH. In our context, where we have a uniform MATH such that each geodesic in MATH is MATH - cobounded, it follows that there is a uniform MATH such that each geodesic in MATH satisfies the MATH contracting projection property. Now consider MATH a geodesic in the NAME graph MATH, mapping via MATH to a piecewise geodesic MATH in MATH, with each subsegment MATH an element of MATH. It follows that MATH is a MATH quasigeodesic in MATH, for MATH independent of the given geodesic in MATH. The MATH - geodesic MATH is MATH - cobounded. Applying REF it follows that MATH, where MATH depends only on MATH. As noted above, closest point projection from MATH onto MATH satisfies the MATH contracting projection property. From this it follows that closest point projection MATH satisfies the MATH contraction property where MATH depend only on MATH. Now define the projection MATH to be the composition MATH where the last map is closest point projection in MATH. This composition clearly satisfies the MATH projection property where MATH depend only on MATH and the quasi-isometry constants and coarse inverse constants for MATH. Geodesics in MATH are clearly coarsely transitive, and applying REF it follows that MATH is word hyperbolic. This means that geodesic triangles in MATH are uniformly thin, and it implies that for each MATH there is a MATH such that MATH quasigeodesic triangles in MATH are MATH - thin. Applying the quasi-isometry between MATH and MATH, it follows that there is a uniform MATH such that for each MATH the geodesic triangle MATH in MATH is MATH - thin; we fix this MATH for the arguments below. Now we turn to a description of the ``limit set" MATH of MATH, with the ultimate goal of identifying it with the NAME boundary MATH. Each geodesic ray in MATH has the form MATH, for some MATH, MATH; define MATH be the set of all such points MATH, over all geodesic rays in MATH. The set MATH is evidently MATH - equivariant. CASE: For any MATH, MATH, the ray MATH in MATH is an element of MATH. To prove this, by definition of MATH there exists a ray MATH in MATH for some MATH. Choose a sequence MATH staying uniformly close to MATH and going to infinity. Pass to a subsequence so that the sequence of segments MATH converges to some ray MATH; it suffices to show that MATH. Since MATH is fixed and the points MATH stay uniformly close to MATH, it follows by REF that the segments MATH stay uniformly close to MATH, and so MATH is in a finite neighborhood of MATH. The reverse inclusion, that MATH is in a finite neighborhood of MATH, is a standard argument: as points move to infinity in MATH taking bounded steps, uniformly nearby points move to infinity in MATH also taking bounded steps, and thus must come uniformly close to an arbitrary point of MATH. This shows that the rays MATH, MATH have finite NAME distance, and applying REF (End Uniqueness) shows that MATH. Note that in the proof of REF we have established a little more, namely that for any MATH and MATH the rays MATH and MATH have finite NAME distance. This will be useful below. CASE: For any MATH there exists a line MATH contained in MATH. From REF it immediately follows that MATH, that the weak hull MATH of MATH is defined, and that MATH acts coboundedly on MATH, since MATH acts coboundedly on MATH. To prove REF , pick a point MATH, and note that by REF we have two rays MATH, MATH in MATH. Pick a sequence MATH staying uniformly close to MATH and going to infinity, and a sequence MATH staying uniformly close to MATH and going to infinity. We have a sequence of triangles MATH in MATH, all MATH - thin. Applying REF there is a MATH such that the sides MATH are contained in the MATH - neighborhood of MATH, and the sides MATH are contained in the MATH - neighborhood of MATH. Each side MATH, being contained in the MATH - neighborhood of MATH, is therefore contained in the MATH - neighborhood of MATH. We claim that the point MATH is uniformly close to the segments MATH. If not, then from uniform thinness of the triangles MATH it follows that there are points MATH and MATH such that the segments MATH and MATH get arbitrarily long while the NAME distance between them stays uniformly bounded. This implies that there are sequences MATH going to infinity and MATH going to infinity such that the NAME distance between the segments MATH and MATH stays uniformly bounded, which implies in turn that the rays MATH and MATH have finite NAME distance. Applying End Uniqueness REF, it follows that MATH, contradicting the hypothesis of REF , and the claim follows. Passing to a subsequence and applying NAME - NAME it follows that MATH converges to a line in MATH. One ray of this line is NAME close to MATH and so has endpoint MATH, and the other ray is NAME close to MATH and so has endpoint MATH, by End Uniqueness. We therefore have MATH, completing the proof of REF . Now we define a map MATH. Recall that the relation of finite NAME distance is an equivalence relation on geodesic rays in the NAME graph MATH of MATH, and MATH is the set of equivalence classes. Consider then a point MATH represented by two geodesic rays MATH and MATH with finite NAME distance in MATH. These map to piecewise geodesic, quasigeodesic rays MATH and MATH with finite NAME distance in MATH. The sequence of geodesic segments MATH in MATH has a subsequence converging to some ray MATH in MATH, and MATH has a subsequence converging to some ray MATH in MATH. To obtain a well defined map MATH it suffices to prove that MATH, and then we can set MATH. To prove that MATH it suffices, by End Uniqueness REF, to prove that the rays MATH and MATH have finite NAME distance in MATH. Since the piecewise geodesic rays MATH have finite NAME distance in MATH, it suffices to prove that MATH has finite NAME distance from MATH, and similarly MATH has finite NAME distance from MATH. Consider a point MATH. For sufficiently large MATH we have MATH. Applying REF there is a uniform constant MATH such that MATH, and so MATH is within distance MATH of some point in MATH. Since MATH is the pointwise limit of MATH as MATH it follows that MATH is within a uniformly bounded distance of MATH. This shows that MATH is within a finite neighborhood of MATH. The reverse inclusion is a standard argument: as points move along MATH towards the end taking bounded steps, uniformly nearby points move along MATH towards the end also taking bounded steps, and thus must come uniformly close to some point of MATH. Hence MATH is well defined. Observe that a similar argument proves a little more: if MATH converges to MATH then the segments MATH converge in the compact - open topology to the ray MATH; details are left to the reader. We now turn to verifying required properties of MATH. To see that MATH is surjective, consider a point MATH and pick a ray MATH in MATH. It follows that MATH is a quasigeodesic ray in MATH. Since MATH is MATH - hyperbolic it follows that MATH has finite NAME distance from some geodesic ray MATH in MATH, with endpoint MATH. As shown above, MATH has finite NAME distance from some geodesic ray MATH. Since MATH are coarse inverses it follows that MATH has finite NAME distance from MATH, and so by End Uniqueness it follows that MATH. To see that MATH is injective, consider two points MATH and suppose that MATH; let MATH be this point. Pick rays MATH in MATH representing MATH respectively. As we have just seen, the images MATH, MATH have finite NAME distance in MATH to rays MATH, MATH in MATH, respectively. As noted at the end of the proof of REF , the rays MATH and MATH have finite NAME distance in MATH; applying the coarse inverse MATH it follows that MATH have finite NAME distance in MATH and therefore MATH. We have shown that MATH is a bijection between MATH and MATH. We want to prove that MATH is a homeomorphism, and that the extension MATH is continuous. For this purpose first we establish: CASE: MATH is a closed subset of MATH, and therefore compact. To prove this, choose a sequence MATH so that MATH in MATH; we must prove that MATH. Choose a point MATH, and apply REF to obtain rays MATH. Passing to a subsequence these converge to a limiting ray MATH in MATH, and so MATH. Looking in the unit tangent bundle of MATH at the point MATH it follows that MATH, and so MATH. CASE: MATH is a homeomorphism. Since both the domain and range are compact NAME spaces it suffices to prove continuity in one direction. Continuity of MATH follows by simply noting that for fixed MATH and for a convergent sequence MATH in MATH, the sequence of rays MATH converges in the compact open topology to the ray MATH. CASE: The map MATH is continuous. To be precise, this map is continuous using the NAME compactification MATH of MATH. We prove this by showing first that the map is continuous using the NAME, and then we apply NAME 's Two Boundaries Theorem CITE which says that the map from the NAME compactification to the NAME compactification is continuous at uniquely ergodic points of MATH. First we recall the NAME compactification in a form convenient for our current purposes. There are actually many different NAME compactifications, one for each choice of a base point in MATH; we shall fix a base point MATH for some MATH. As we have seen, there is a unique geodesic segment MATH for each MATH, and a unique geodesic ray MATH for each MATH. The NAME topology on MATH restricts to the standard topologies on MATH and on MATH, it has MATH as a dense open subset, and a sequence MATH converges to MATH if and only if the sequence of segments MATH converges to the ray MATH in the compact open topology; equivalently, letting MATH denote the unit ball in MATH centered on MATH, the distance MATH goes to infinity and the set MATH converges to the set MATH in the NAME topology. We already proved in REF that MATH is continuous; for this we implicitly used the fact that the NAME topology on MATH is identical to the NAME topology, defined by identifying MATH with the unit tangent bundle at MATH. We also observed earlier, after the proof that MATH is well-defined, that if MATH converges to MATH, then MATH converges to MATH in the NAME topology on MATH. Putting these together it follows that MATH is continuous using the NAME topology on MATH. Since MATH consists entirely of uniquely ergodic points in MATH, NAME 's Two Boundaries Theorem CITE implies that the identity map on MATH is continuous from the NAME topology to the NAME topology at each point of MATH, and so MATH is continuous using the NAME topology on MATH. We now put the pieces together to complete the proof of convex cocompactness. Let MATH be an arbitrary MATH - equivariant map, and define MATH to be equal to MATH. We must prove that MATH is a quasi-isometry and that the extension MATH is continuous. From REF - REF above, it follows that the quasi-isometry MATH has continuous extension MATH, and so MATH is a NAME hyperbolic metric space with NAME compactification MATH. Since MATH is a MATH - invariant subset, it follows that MATH is NAME hyperbolic with NAME compactification MATH. The map MATH is a MATH - equivariant map between quasigeodesic metric spaces on which MATH acts properly and coboundedly by isometries, and hence MATH is a quasi-isometry. Since MATH is uniformly bounded for MATH, then from the fact that MATH it follows that MATH is continuous. This completes the proof that weak orbit quasiconvexity implies convex cocompactness. CASE: Assuming MATH is convex cocompact, pick a finite generating set for MATH with NAME graph MATH and MATH - equivariant, coarsely inverse quasi-isometries MATH, MATH. Let MATH be an orbit of MATH in MATH. Since MATH acts coboundedly on MATH it follows that MATH has finite NAME distance from MATH in MATH. It suffices to show that for any two points MATH there is a geodesic line whose infinite ends are in MATH such that MATH come within a uniformly finite distance of that line. Pick a MATH - equivariant map MATH taking the vertices of MATH bijectively to MATH and each edge of MATH to a geodesic segment, so MATH and MATH differ by a bounded amount. Since MATH is MATH - hyperbolic it follows that there is a constant MATH such that any two vertices of MATH lie within distance MATH of some bi-infinite geodesic. Pick MATH, and pick a bi-infinite geodesic MATH in MATH such that MATH are within distance MATH of MATH. Let MATH be the two ends of MATH. By the statement of convex cocompactness, there is a MATH quasigeodesic line in MATH of the form MATH whose two infinite ends are MATH, where MATH are independent of MATH. It follows that MATH and MATH are uniformly close, and so MATH and MATH are uniformly close, and so the points MATH are uniformly close to MATH. |
math/0106190 | A standard lemma found in most O.D.E. textbooks shows that if MATH is a smooth flow on a compact manifold then there is a constant MATH such that MATH. We can plug into this argument as follows. The conclusion of the lemma is local, and so it suffices to prove it under the assumption that MATH and that MATH is affine. There exists a compact subset MATH such that any MATH - cobounded, MATH - lipschitz path MATH, can be translated by the action of MATH to lie in the set MATH. Let MATH be the set of all MATH - lipschitz affine paths MATH, a compact space in the compact open topology. By naturality of the metric on MATH, it suffices to prove the lemma for MATH. For each MATH and each vector MATH tangent to a fiber MATH, MATH, define: MATH . Since MATH for MATH, we may regard MATH as a function defined on the projective tangent bundle of MATH crossed with MATH, a compact space. As MATH varies, and as MATH varies over the compact space MATH, the function MATH varies continuously, and so by compactness MATH has a finite upper bound MATH. Setting MATH, it now follows by standard methods that MATH when MATH is tangent to MATH, and so MATH is MATH bilipschitz. |
math/0106190 | Both REF are proved in the same manner using REF ; we prove only REF . To smooth the notation in the proof we denote MATH, we let MATH denote the fiber MATH of MATH, we let MATH denote the corresponding fiber MATH of MATH, etc. To prove REF , by applying REF we choose for each MATH a marked map MATH for which any lift MATH is a MATH quasi-isometry, where the constants MATH depend only on MATH. Since each MATH preserves markings we may choose the lifts MATH so that for any MATH we have a commutative diagram of induced boundary maps: MATH . Applying REF it follows that if we strip off the MATH symbols from the above diagram, and if we choose MATH so that MATH, then we obtain the following diagram, a coarsely commutative diagram in the sense that the two paths around the diagram differ in the sup norm by a constant MATH depending only on MATH: MATH . Define MATH so that MATH. To prove that MATH is a quasi-isometry we need only show that if MATH satisfy MATH then MATH is bounded by a constant depending only on MATH, and then carry out the similar argument with inverses. Given MATH with MATH, choose MATH so that MATH, MATH. By the metric fibration property we have MATH. Changing notation if necessary we may assume that MATH. Let MATH be the geodesic in MATH connecting MATH and MATH, and by the metric fibration property note that MATH. Consider the map MATH whose restriction to MATH is the connection map MATH; it follows that MATH is bilipschitz with constant MATH. The distance in MATH between the point MATH and the point MATH is therefore at most MATH. Mapping over to MATH we have MATH and since MATH, the proof is done. |
math/0106190 | This follows because the subgroup of MATH stabilizing MATH is the normal subgroup MATH, and the inclusion map MATH is uniformly proper with respect to word metrics, a fact that holds for any finitely generated subgroup of a finitely generated group. |
math/0106190 | We use MATH for the metric on MATH. First observe that any MATH - quasivertical path MATH in MATH is a MATH - quasigeodesic in MATH, in fact MATH . The upper bound is just the fact that MATH is MATH - lipschitz, and the lower bound follows from the metric fibration property for MATH, together with the fact that MATH is a geodesic in MATH. Consider then a pair of MATH - quasivertical paths MATH defined on a subinterval MATH, and let MATH. We assume that MATH is even and let MATH. For each MATH we have a fiber MATH isometric to MATH, with metric denoted MATH. We must prove that the sequence MATH satisfies MATH flaring, with MATH independent of MATH and with MATH independent of MATH, MATH, and MATH. For MATH let MATH be the connection map, a MATH bilipschitz map. For each MATH we have a MATH geodesic MATH with endpoints MATH, MATH. There is a family of quasivertical paths MATH described as follows: CASE: For each MATH and each MATH the family contains a unique quasivertical path MATH that passes through the point MATH. If we fix MATH, we thus obtain a parameterization of the family MATH by points MATH. CASE: The ordering of the family MATH induced by the order on MATH is independent of MATH. The first path MATH in the family is identified with MATH, and the last path MATH is identified with MATH. CASE: Each MATH is MATH - quasivertical, where MATH depends only on MATH and MATH. When MATH is assumed fixed, we write MATH for the path MATH. Given MATH, consider the following MATH - quasigeodesic in MATH: MATH . Since connection paths are geodesics, and since MATH are MATH - quasivertical, it follows that the endpoint MATH and the corresponding endpoint MATH have distance in MATH at most MATH, and similarly for the opposite endpoints MATH and MATH. Each endpoint of MATH and the corresponding endpoint of MATH therefore have distance in MATH bounded by a constant depending only on MATH; this follows from REF . Since the spaces MATH are all isometric to MATH, it follows that the NAME distance between MATH and MATH in MATH is bounded by a constant depending only on MATH, MATH, which implies in turn that there is a quasi-isometric reparameterization MATH such that MATH where the constant MATH and the quasi-isometry constants for MATH depend only on MATH, MATH. By possibly increasing the quasi-isometry constants we may assume furthermore that MATH is an orientation preserving homeomorphism. It follows that we may connect the point MATH to the point MATH by a MATH - quasivertical path defined over the interval MATH, where MATH depends only on MATH, MATH; when MATH we may choose the path to be MATH, and when MATH we may choose the path MATH. By piecing together these paths as MATH varies over MATH, we obtain the required family of paths MATH. We use MATH - hyperbolicity of MATH in the following manner. First, for any geodesic rectangle MATH in MATH it follows that any point on MATH is within distance MATH of MATH. Second, for any MATH quasigeodesic in MATH, the NAME distance to any geodesic with the same endpoints is bounded by a constant MATH depending only on MATH. For any rectangle of the form MATH where MATH are geodesics and MATH are MATH quasigeodesics, it follows that any point on MATH is within distance MATH of MATH. By REF there exists a constant MATH such that: MATH . We are now ready to define the flaring parameters MATH. Let MATH where MATH is the greatest integer MATH. Assuming as we may that MATH (and so the NAME distance between MATH and MATH in MATH equals MATH), we must prove: CASE: if MATH then MATH . CASE: It follows that there is a rectangle in MATH of the form MATH where MATH is a geodesic in MATH with the same endpoints as MATH, and where MATH has length MATH. Consider now the point MATH, whose distance from some point MATH is at most MATH. If MATH then it follows that MATH a contradiction. We reach a similar contradiction if MATH. Therefore MATH. It follows that MATH for some MATH such that MATH, and so by following along MATH a length at most MATH we reach the point MATH. This shows that MATH, and so MATH, that is, MATH. CASE: In the family MATH, we claim that there is a discrete subfamily MATH, with MATH, such that the following property is satisfied: for each MATH, letting MATH then we have MATH . By assumption of REF , the subfamily MATH has the property MATH (for MATH). Suppose by induction that we have a subfamily MATH, with MATH, such that MATH for all MATH, but suppose that MATH for some MATH. If, say, MATH, then we subdivide the geodesic segment MATH in half at a point MATH, yielding two subsegments of length MATH, and we add the path MATH to our subfamily; similarly, if MATH then we subdivide the interval MATH in half. This process must eventually stop, because MATH thereby proving the claim. From the exact same argument as in REF , using the fact that MATH it now follows that MATH for all MATH. We therefore have: MATH . This completes the proof of REF . |
math/0106190 | This is basically an immediate application of the NAME - NAME combination theorem CITE. To be formally correct, some remarks are needed to translate from our present geometric setting, of a hyperbolic plane bundle MATH, to the combinatorial setting of CITE, and to justify that our vertical flaring property for MATH corresponds to the ``hallways flare condition" of CITE. We may assume that the endpoints of the interval MATH, if any, are integers. The first observation is that there is a MATH - equivariant triangulation MATH of MATH with the following properties: CASE: For each MATH there is a REF-dimensional subcomplex MATH which is a triangulation of the hyperbolic plane MATH. CASE: Each REF-cell of MATH is either horizontal (a REF-cell of some MATH), or vertical (connecting a vertex of some MATH to a vertex of some MATH); CASE: each REF-cell of MATH is either horizontal (a REF-cell of some MATH), or vertical (meaning that the boundary contains exactly two vertical REF-cells). CASE: There is an upper bound depending only on MATH, MATH for the valence of each REF-cell and the number of sides of each REF-cell. CASE: The inclusion of REF-skeleton of MATH into MATH is a quasi-isometry with constants depending only on MATH and MATH. To see why MATH exists as described, consider the marked hyperbolic surface bundle MATH. For each hyperbolic surface MATH, MATH, there is a geodesic triangulation MATH of MATH with one vertex, whose edges have length bounded only in terms of MATH. It follows that there are constants MATH, MATH depending only on MATH, such that if MATH is the lifted triangulation in MATH, then the inclusion of REF-skeleton of MATH into MATH is a MATH quasi-isometry. Then, regarding MATH as a triangulation of MATH, we can extend to a cell-decomposition MATH of MATH which is a graph of spaces of bounded combinatorics. The existence of MATH uses the fact that each connection map MATH is MATH - bilipschitz, so by moving each vertex of MATH along a connection path into MATH and them moving a finite distance to a vertex of MATH we obtain a MATH - quasi-isometry MATH, with MATH depending only on MATH, and from this we easily construct MATH so that its lift MATH has the desired properties. The second observation is that vertical flaring in MATH is equivalent to the ``hallway flare condition" of CITE for MATH, and this equivalence is uniform with respect to the parameters in each property. To see why, note that quasivertical paths in MATH correspond to thin paths in MATH as defined implicitly in REF: an edge path MATH is MATH - thin if the restriction of MATH to each subinterval MATH lies in MATH and is a concatenation of at most MATH edges. Under the quasi-isometry MATH and its coarse inverse MATH, MATH - quasivertical paths in MATH correspond to MATH - thin paths with a uniform relation between MATH and MATH. In order to complete the translation from the geometric setting to the combinatorial setting, while the results of CITE are stated only when MATH is the universal cover of a finite graph of spaces, nevertheless, the proofs hold as stated for any graph of spaces with uniformly bounded combinatorics: all the steps in the proof extend to such graphs of spaces, regardless of the presence of a deck transformation group with compact quotient. The conclusion of the combination theorem is the MATH - hyperbolicity of REF-skeleton of MATH, with MATH depending only on the flaring constants for MATH, which depend in turn only on MATH, MATH, and the flaring constants for MATH. It follows that MATH is MATH hyperbolic with the correct dependency for the constant MATH. |
math/0106192 | There is no loss of generality in assuming MATH. CASE: First suppose MATH; without loss of generality, we may assume MATH. Write MATH with MATH and MATH. Set MATH . The first inner sum converges because MATH for MATH, while the second inner sum converges because MATH as MATH runs over any decreasing sequence. It is easily checked that MATH and that MATH. Let MATH be a solution of MATH; then we may set MATH to obtain a solution of MATH. Next, suppose MATH. Write MATH with MATH and MATH. (The index REF could be replaced by any positive index.) We wish to set MATH . The first sum converges MATH-adically in MATH or MATH to a solution of MATH. The second sum converges MATH-adically (since MATH has no coefficients of index less than MATH) to a solution of MATH. Thus we may set MATH to obtain a solution of MATH. CASE: Set MATH the inner sum converges because MATH is bounded and MATH. Then it is easily verified that MATH. CASE: Suppose MATH; put MATH and let MATH be the smallest index such that MATH. First suppose MATH. Comparing the coefficients of MATH in the equation MATH, we have MATH. In this equation, MATH but MATH, so MATH. On the other hand, MATH since MATH, contradiction. Now suppose MATH. In this case, the above argument gives MATH. By comparing the coefficients of MATH as well, we obtain MATH by induction on MATH. But this conclusion contradicts the fact that MATH is bounded for MATH. |
math/0106192 | Let MATH be an element of MATH. (The proof for MATH is analogous, so we omit reference to it hereafter.) Let MATH be an irrational number, so that MATH occurs for a unique value of MATH. Without loss of generality, we may assume that value is MATH and that MATH. (Otherwise, we can multiply by a suitable constant times a suitable power of MATH.) Put MATH then by construction, MATH. Now define a sequence MATH as follows. Let MATH be the smallest index such that MATH. (If no such MATH exists, then MATH and there is nothing to prove.) Set MATH. To define MATH from MATH, set MATH and let MATH . We will show that MATH converges, in a suitable sense, to a limit MATH and that MATH. Suppose MATH satisfies MATH for all MATH. Let MATH, MATH and MATH. Then MATH; if we put MATH, it follows that MATH for all MATH. Then MATH . Therefore for MATH nonzero, MATH . For MATH, the initial inequality holds for MATH. Therefore for MATH, MATH by induction on MATH, and the same inequality holds for the MATH as noted above. Let MATH be the subring of MATH consisting of series MATH such that MATH for all MATH. Then MATH carries a valuation MATH defined by MATH. With respect to MATH, the product MATH converges; we call the limit MATH. Moreover, we can extend MATH to the subring of MATH defined by MATH for all MATH, in which it is clear that MATH. In particular, for MATH, MATH. Thus MATH, and we may factor MATH as MATH with MATH and MATH, as desired. |
math/0106192 | Let MATH be a NAME sequence in the NAME topology, and put MATH. Then by hypothesis, for MATH, MATH and MATH, there exists MATH such that for MATH, MATH. In particular, for each fixed MATH, MATH is a NAME sequence in MATH, so it has a limit MATH. Now put MATH. For any given MATH and MATH, choose MATH so that for MATH, MATH. Then MATH for all MATH. Since the absolute value on MATH is nonarchimedean, we also have MATH, so MATH. We conclude that MATH. |
math/0106192 | Since MATH, MATH is invertible in that ring, and MATH can be written as MATH, where MATH satisfies MATH for MATH. Now write MATH, and set MATH. Then MATH has no coefficients of positive index, and so MATH has no coefficients of index MATH or greater. Note that MATH is well-defined; since MATH, we must have MATH. Thus MATH. Replacing some coefficients of a series with zeroes cannot increase its norm, so MATH and MATH. |
math/0106192 | Fix a slope MATH. Suppose the NAME polygon of MATH intersects its support line of slope MATH from MATH to MATH, and the NAME polygon of MATH intersects its support line of slope MATH from MATH to MATH, with MATH and MATH. We claim the NAME polygon of MATH intersects its support line of slope MATH at MATH and MATH, which would imply the statement of the lemma. To verify the claim, we first note that MATH with equality if the minimum occurs for a single value of MATH. For MATH, we have MATH and MATH, so MATH. Similarly, if MATH, we have MATH and MATH, so again MATH. Thus the minimum is achieved only for MATH, and so MATH. Likewise, MATH. By a similar argument, we also have that for MATH, MATH and MATH. Namely, MATH the second inequality is strict for MATH, and the third is strict for MATH, so MATH and the inequality follows; the other inequality follows analogously. (The minimum really is a minimum, not an infimum, so termwise strict inequality implies strict inequality for the minimum.) Thus the NAME polygon of MATH intersects its support line of slope MATH from MATH to MATH, respectively. |
math/0106192 | We construct a sequence of elements MATH of MATH supported on MATH, having constant coefficient REF, convergent under the norm MATH, such that MATH; this will imply that MATH and MATH converge in the NAME topology as well, and that if MATH, then MATH. Specifically, we set MATH . Let MATH; by construction, MATH. We prove by induction that MATH. This holds by design for MATH. Now suppose it holds up to some MATH. Then we can write MATH with MATH, and MATH . Since MATH by the induction hypothesis, and MATH, we have MATH. Thus MATH . We conclude that MATH, completing the induction. By the induction, we have that MATH is NAME, hence convergent, and that MATH converges to REF. Thus the limit MATH of MATH satisfies MATH, as desired. |
math/0106192 | Let MATH. Without loss of generality, assume MATH, and let MATH be the smallest index such that MATH. Let MATH; if we put MATH, then MATH for MATH. That is, MATH, so MATH is the unit ideal, as then is MATH. |
math/0106192 | CASE: Without loss of generality, we may assume MATH is truncated and MATH has constant coefficient REF. Put MATH; then MATH, and MATH. Thus if we put MATH, we also have MATH, so the series MATH converges in the NAME topology to a limit MATH. Likewise, MATH converges to REF, so MATH is divisible by MATH, and MATH is thus the unit ideal. CASE: Let MATH and MATH be the multiplicities of MATH as a slope of MATH and MATH. Then MATH and MATH are integral multiples of MATH, where MATH is the smallest positive valuation of MATH. Thus we can induct on MATH. Assume without loss of generality that MATH, and that MATH is truncated. Let MATH, so that MATH and it suffices to show that MATH is generated by a pure element of slope MATH. If MATH, then MATH and we are done, so assume MATH. Otherwise, let MATH and choose MATH to minimize MATH. Since MATH is the unit ideal by the previous lemma, we can find MATH such that MATH, and MATH. Let MATH; then MATH and all slopes of MATH are at least MATH. Moreover, the multiplicity of MATH as a slope of MATH is strictly less than MATH. By REF , we can factor MATH as MATH with MATH pure of slope MATH with the same multiplicity, and MATH having all slopes greater than MATH. By REF , MATH is the unit ideal, so MATH, which is equal to MATH, is also equal to MATH. By the induction hypothesis, MATH is generated by a pure element of slope MATH, as desired. |
math/0106192 | If MATH divides MATH, then obviously any factor of MATH divides MATH. Conversely, suppose MATH is a slope factorization of MATH. Put MATH; then the sequence MATH is NAME, so has a limit MATH, and clearly MATH. |
math/0106192 | Let MATH be the slope of MATH, and let MATH be the smallest positive valuation of MATH. Put MATH; then MATH for MATH. In particular, MATH is a unit in MATH; let MATH be its inverse and put MATH. If we put MATH, then MATH for MATH by construction and MATH for MATH. In particular, we have MATH for MATH. Therefore for any fixed MATH, eventually MATH; since MATH, MATH. We conclude that MATH converges in the NAME topology, and we may take the limit as our desired MATH. |
math/0106192 | Without loss of generality, assume MATH has nonzero constant coefficient. Factor MATH as MATH as in REF , so that MATH is pure of slope equal to the first slope of MATH, with the same multiplicities. Then by REF , the NAME polygon of MATH is equal to that of MATH with its first segment removed. Then factor MATH as MATH in the same fashion, and so on. If the NAME polygon of MATH has finitely many slopes, then this process eventually represents MATH as a product of pure elements, as desired. If the NAME polygon of MATH is infinite, REF allows us to construct an element MATH with a slope factorization whose slope factors are unit multiples of the MATH. In particular, MATH and MATH have the same NAME polygon. Since MATH is divisible by each MATH, by REF MATH is divisible by MATH. Moreover, the NAME polygon of MATH is empty, so MATH is a unit. We can modify the slope factorization of MATH by multiplying its first factor by MATH to obtain the desired slope factorization of MATH. |
math/0106192 | Let MATH be the slope of MATH, and let MATH be the smallest positive valuation in MATH. By imitating the proof of REF , we may assume that MATH is such that MATH and MATH for MATH. In particular, this implies that MATH converges to a limit MATH; let MATH. We construct a sequence MATH such that MATH and MATH converges; then we may set MATH and be done. First, choose MATH with MATH, which exists because MATH and MATH are relatively prime. It suffices to show that for any MATH, there exists MATH such that for all MATH, MATH; for example, one can then choose such a MATH with MATH for all MATH, and set MATH. Note that MATH for MATH, so in those norms, the sequence MATH is NAME. In particular, there exists MATH such that MATH for all MATH. Set MATH; it is now clear that the series MATH is NAME for all of the MATH, and thus convergent. We then take MATH and the proof is complete. |
math/0106192 | The constant coefficient and power of MATH are clearly uniquely determined by MATH, so we may suppose MATH and MATH are slope factorizations of MATH. It suffices to prove that each MATH divides one of the MATH (and vice versa). Taking the greatest common divisor of MATH with the MATH of the same slope, if it exists, and dividing out that divisor, we may reduce to the case where each MATH is relatively prime to MATH. In this case, for each MATH, there exists MATH such that MATH. By REF , there exists MATH such that MATH for each MATH, so by REF MATH is divisible by MATH. But then MATH and MATH are both divisible by MATH, a contradiction since MATH is not a unit. Thus MATH divides one of the MATH, as desired. |
math/0106192 | It suffices to show that for MATH (respectively, MATH), there exist MATH such that MATH generates the ideal MATH. By REF , we may reduce to the case where MATH (respectively, MATH) and have nonzero constant coefficients. By REF , there exists MATH whose slope factorization consists of the greatest common divisors of the slope factors of MATH and MATH, and by REF , MATH and MATH are divisible by MATH. Dividing off MATH, we reduce to the case where MATH have no common slope factors. Let MATH be the slope factorization of MATH. By assumption, MATH is not divisible by MATH, so there exists MATH such that MATH. By REF , there exists MATH such that MATH; then MATH is divisible by all of the MATH. Therefore there exists MATH such that MATH, and MATH as desired. |
math/0106192 | Observe that for MATH, MATH . Thus we can modify MATH by a suitable linear combination of MATH to obtain MATH such that MATH for some MATH. Suppose MATH satisfies MATH. Comparing coefficients in this equation, we have MATH, MATH for MATH, and MATH. If MATH and MATH satisfy the equations MATH then setting MATH and MATH for MATH and MATH satisfies MATH. Thus it suffices to show that REF has a solution. Notice that replacing MATH by MATH does not alter whether a solution exists: for any MATH such that MATH, MATH . If we write MATH with MATH and MATH, then the equation MATH has a solution by REF . Thus without loss of generality, we may assume MATH for MATH. In particular, we now have MATH. We next reduce to the case where MATH is supported on MATH. Set MATH . The first sum is evidently convergent in MATH. As for the second, recall that there exists constants MATH such that MATH for MATH. Thus MATH and the right side grows exponentially in MATH, so the second sum is also convergent. Thus we can replace MATH by MATH which is supported on MATH. We now assume MATH is supported on MATH. If MATH, then MATH has an eigenvector MATH with MATH, which can be multiplied by a suitable scalar to produce the desired MATH, so we assume MATH; indeed, we may assume MATH. Define MATH for all MATH by setting MATH for MATH and extending by the rule MATH. We say an index MATH is a corner of MATH if MATH for all MATH; then all corners are of the form MATH for MATH in a finite set and MATH an arbitrary integer. The solutions of MATH have the form MATH for MATH. For such MATH, define MATH . Then MATH is additive, and MATH if and only if the equation MATH has a solution. For MATH and MATH, put MATH. Write MATH . Let MATH be the smallest MATH which achieves MATH for some MATH, let MATH be the smallest such MATH, and let MATH. Then MATH is a corner of MATH if MATH. On the other hand, if MATH and MATH were not a corner, we could find MATH with MATH. Let MATH be the unique integer such that MATH. Then MATH contradiction. Thus MATH is also a corner if MATH, so MATH is piecewise constant, as is MATH; therefore MATH is piecewise linear and increasing in MATH. Also, clearly MATH and MATH. We claim that if MATH and MATH lie in MATH for some MATH, then MATH is continuous at MATH. Let MATH and MATH be the value of MATH and MATH, respectively, for small MATH; define MATH and MATH analogously. If MATH, then MATH for MATH, contradicting the definition of MATH. A similar contradiction arises if MATH. Hence MATH and MATH have the same norm. Now by the definition of MATH, we have MATH and MATH. Thus MATH and MATH is continuous at MATH. We can also show that MATH if and only if MATH, but the argument is more delicate. Define MATH as above. If MATH but MATH, then for MATH we have MATH. On the other hand, for MATH, we can take MATH and obtain MATH . Thus MATH is sandwiched between MATH and MATH, but there is no norm between these two because MATH is a uniformizer, contradiction. If MATH but MATH, then MATH. On the other hand, for MATH, we can take MATH and obtain MATH but MATH, so again we obtain a contradiction. Since MATH and MATH is increasing and continuous except for jumps between MATH and MATH, it follows that MATH maps MATH onto MATH one or more times. Choose MATH such that MATH; then MATH also maps MATH onto MATH one or more times. It follows that for any MATH supported on MATH with MATH, there exists MATH with MATH with MATH for MATH such that MATH. This MATH can be constructed by a transfinite recursion: find MATH and MATH such that MATH has leading coefficient congruent to the leading coefficient of MATH modulo MATH, then subtract off and repeat. Additionally, note that MATH must change slope at some point in MATH (possibly equal to MATH), since MATH. If MATH changes slope at MATH, then MATH has at least two distinct terms with minimal norm. Namely, if again MATH and MATH (respectively, MATH and MATH) are the values of MATH and MATH (respectively, MATH and MATH) for MATH small, then the terms MATH and MATH have the same minimal norm. Note that there exists MATH such that the sum of the terms of minimal norm has norm less than REF (because the residue field of MATH is algebraically closed). Thus in the transfinite recursion of the previous paragraph, there is more than one choice that can be made at MATH. In particular, there exists MATH with MATH, MATH and MATH for some MATH. From this analysis, we can construct a nonzero solution of REF . Start with MATH such that MATH, MATH and MATH for some MATH. Now construct a sequence MATH such that CASE: MATH for all MATH; CASE: MATH; CASE: MATH for MATH. Specifically, given MATH, let MATH, find MATH such that MATH with MATH for MATH and MATH, then set MATH. Then MATH converges in the NAME topology to a nonzero MATH with MATH and MATH. Thus REF has a nonzero solution, and the proof is complete. |
math/0106192 | We imitate the previous proof restricted to MATH. For starters, given MATH supported on MATH with MATH, it again suffices to show that the equations MATH have a solution with MATH not both zero (the case MATH is self-evident). Define MATH for all MATH by setting MATH for MATH and extending by the rule MATH. Again, we say an index MATH is a corner of MATH if MATH for all MATH. For MATH a solution of MATH, define MATH it suffices to exhibit MATH such that MATH. Continuing to imitate the previous proof, for MATH, define MATH, MATH, MATH by taking MATH as the smallest MATH for which MATH is achieved, MATH as the smallest MATH for which the maximum is achieved with MATH, and MATH. Then as before, MATH and MATH are piecewise constant and so MATH is increasing and piecewise linear. We again prove that MATH and MATH are either equal, or equal to MATH and MATH, respectively. The proof that if both limits lie in MATH, then they are equal, carries through as before, as does the proof that we cannot have MATH and MATH. Now suppose MATH but MATH. For MATH we have MATH. On the other hand, for MATH, we can take MATH and obtain MATH . Thus MATH is sandwiched between MATH and MATH, but there is no norm between these two because MATH is a uniformizer, contradiction. Given the results of the previous paragraph, the rest of the proof proceeds as in the previous lemma. Namely, a transfinite recursion can be used to generate a solution of MATH, which completes the proof. |
math/0106192 | Since MATH has nonzero determinant, we can find a permutation matrix MATH such that MATH has nonzero diagonal entries. Put MATH and MATH; then MATH is diagonal and its entries are a permutation of those of MATH. From MATH we have MATH for each MATH; since MATH is nonzero, this implies MATH. In particular, the MATH-th largest slope of MATH is greater than or equal to the MATH-th largest slope of MATH for each MATH. The analogous statement with MATH and MATH reversed follows by a similar argument; therefore the slopes of MATH and MATH are equal up to permutation. |
math/0106192 | Let MATH be the coordinate ideal of MATH. By REF , MATH is principal, so we may choose a generator MATH. Since MATH, the ideal MATH is invariant under MATH and MATH, so MATH for MATH a unit in MATH. By REF we can write MATH with MATH and MATH a unit in MATH. The equation MATH has a solution with MATH, and MATH satisfies MATH and generates MATH. Therefore there exists MATH with MATH, MATH primitive, and MATH, as desired. |
math/0106192 | We proceed by induction on MATH. For MATH, MATH acts on a basis vector by an invertible scalar, that is .an element of MATH. Without loss of generality, we may assume this scalar is in MATH and has norm REF, in which case the result follows from CITE. Now suppose MATH. We are done if MATH is isomorphic to MATH, so we assume that this does not occur. Let MATH be the slope of MATH. For MATH a rational number, define the MATH-index of MATH as the smallest positive integer MATH such that MATH is a valuation of MATH. Since the set of rationals of MATH-index less than MATH is discrete, there is a smallest such rational that occurs as the slope of an eigenvector of MATH over a suitable extension of MATH; call this number MATH. Let MATH be the MATH-index of MATH, and let MATH have valuation MATH. Let MATH be an eigenvector of MATH over MATH with MATH. Write MATH, so that each MATH is an element of MATH over MATH with MATH, and let MATH be the span of MATH within MATH. Since MATH, we may apply the induction hypothesis to MATH. If MATH has more than one standard summand, it has an eigenvector of slope less than or equal to MATH with MATH-index strictly less than MATH. Thus MATH has an eigenvector of slope strictly less than MATH with MATH-index less than MATH, contradiction. Thus MATH itself is standard; specifically, it must be isomorphic to MATH (in particular, MATH must equal MATH). Likewise, MATH has a direct sum decomposition MATH of the specified type. Let MATH be the preimage of MATH in MATH; to complete the proof, it suffices to show that the exact sequence MATH splits for MATH. First suppose MATH. Then the dimension of each MATH is less than MATH, so we can apply the induction hypothesis to MATH. If the slope of MATH were less than that of MATH, then the induction hypothesis would imply that MATH has a slope less than MATH of MATH-index less than or equal to MATH, contradicting the minimality of MATH. Thus the slope of MATH is greater than or equal to that of MATH. In this case, REF can be used to show that the exact sequence splits. Namely, by imitating the argument at the beginning of the proof of REF , we can reduce this splitting to the existence of solutions of equations of the form MATH, where MATH is the least common multiple of MATH and MATH, and MATH. Then REF implies that each of these equations has a solution. The case MATH requires special scrutiny, as MATH and the induction hypothesis does not apply. Let MATH be the slope of MATH. As above, the exact sequence splits if MATH, so assume on the contrary that MATH; this implies in particular that MATH. If MATH, we immediately obtain a contradiction from REF , so we may assume MATH. We show that MATH has an eigenvector of slope less than or equal to MATH over some finite extension of MATH. Pick an eigenvector MATH of slope MATH, and apply the induction hypothesis to the quotient of MATH by the span of this eigenvector. This yields an eigenvector MATH of the quotient of slope at most MATH. Applying the induction hypothesis again, this time to the preimage of MATH, gives an eigenvector of MATH of some slope MATH. Let MATH be a finite extension of MATH whose value group contains MATH, and let MATH be the smallest rational of MATH-index less than MATH that occurs as the slope of an eigenvector. Again, take the quotient of MATH by the span of an eigenvector of slope MATH, this time over MATH and apply the induction hypothesis. If its slopes are not all equal, we deduce that MATH has the desired splitting over MATH, and in particular has an eigenvector of slope less than or equal to MATH over MATH. Otherwise, we can repeat the process to produce an extension MATH of MATH, and the smallest rational MATH of MATH-index less than MATH occuring as the slope of an eigenvector will be at most MATH, and so on. The existence of an eigenvector of slope at most MATH is assured if the above process ever terminates, so assume it continues indefinitely. Then the sequence MATH converges to MATH, as it is sandwiched between MATH and a sequence obtained by iterating MATH, and the latter converges to MATH; therefore, there exist eigenvectors of MATH of every rational slope greater than MATH. Let MATH be an extension of MATH whose value group contains MATH, and let MATH be its minimum positive valuation. Take an eigenvector MATH of MATH of slope MATH over an extension of MATH of degree MATH. The span of MATH over MATH has dimension at most MATH and sum of slopes at most MATH. Apply the induction hypothesis to the span; if the sum of slopes is not equal to MATH, then it is at most MATH, and the span contains an eigenvector of slope at most MATH. If the sum of slopes is equal to MATH but the slopes are not all equal, and MATH is the least slope and its multiplicity is MATH, then MATH; since MATH and MATH are integral multiples of MATH, we deduce MATH and the span again contains an eigenvector of slope at most MATH. Finally, if all of the slopes of the span are equal to MATH, then REF implies that MATH has an eigenvector of slope MATH. Thus in all cases, MATH has an eigenvector of slope less than or equal to MATH. Let MATH be an element of a finite extension of MATH of valuation MATH. Because MATH has an eigenvector of slope less than or equal to MATH, it must also have one of slope equal to MATH, over some finite extension of MATH. In fact, from a solution of MATH over a finite extension of MATH, we can obtain a solution over MATH: decompose the solution over a basis of the finite extension over MATH, and choose any nonzero component. Thus we may assume MATH is defined over MATH. Let MATH be the MATH-index of MATH; then we can write MATH. Let MATH be the span of MATH. If MATH, then we can apply the induction hypothesis to MATH to express it as a direct sum of standard subcrystals. The projection of MATH onto at least one of these subcrystals must be nonzero, and since MATH, the same equation holds for the projection of MATH. Thus this subcrystal has an eigenvector of slope MATH; its slope must then be less than MATH. Since this slope has MATH-index at most MATH, we conclude MATH, contradiction. On the other hand, if MATH, then MATH is isomorphic to MATH. Moreover, MATH is an eigenvector of MATH of slope MATH, so must be primitive. Thus MATH is isomorphic to MATH, contrary to an earlier assumption. We conclude that the assumption MATH leads to a contradiction in all cases, so we must have MATH and so the exact sequence MATH splits. In summary, we have that each MATH splits as a direct sum of MATH with another summand; now MATH splits as a direct sum of these other summands with MATH, as desired. |
math/0106192 | If MATH form a basis of eigenvectors of MATH with MATH, then MATH so MATH is an eigenvector of slope MATH. |
math/0106192 | By the previous proposition, it suffices to prove that the highest special slope is no greater than the highest generic slope. Let MATH be a basis of eigenvectors of MATH and let MATH be an eigenvector of MATH of highest slope over MATH, which exists by CITE. Write MATH with MATH. If MATH and MATH, then for MATH, we have MATH, so MATH. The latter equation only has solutions in MATH if MATH. Thus each special slope is less than or equal to the highest generic slope. |
math/0106192 | This follows immediately from REF . |
math/0106192 | It suffices to show that the eigenvectors of MATH of lowest slope are defined over MATH. Let MATH be eigenvectors of MATH over MATH, with MATH for MATH such that MATH. By the descending slope filtration for MATH-crystals CITE, we can find elements MATH of MATH over MATH such that MATH for some MATH. Write MATH with MATH; then applying MATH to both sides, we have MATH for MATH. For MATH, this implies MATH. By REF and descending induction on MATH, we obtain MATH for MATH, and so MATH is defined over MATH. On the other hand, by the ascending slope filtration CITE, the eigenvectors MATH of MATH over MATH satisfying MATH are defined over MATH. Thus MATH is defined over MATH. Since MATH could have been taken to be any eigenvector over MATH of lowest slope, this proves the claim. |
math/0106193 | Let MATH be an element of MATH whose reduction modulo MATH generates MATH over MATH. Then for MATH sufficiently large, MATH is a MATH-semiunit. Thus if we take MATH, we may use powers of MATH as a MATH-semiunit basis. |
math/0106193 | Factor MATH as a product of elementary matrices in MATH. It suffices to note that for each elementary matrix MATH in this product and each MATH, there exists MATH over MATH for some finite extension MATH of MATH such that MATH and MATH. This is vacuous for permutation matrices and easily verified for the other two kinds of matrices. |
math/0106193 | Fix MATH such that MATH and MATH are defined, and let MATH be the smallest positive valuation of MATH and MATH. By REF , for any MATH, we can change basis over MATH for some finite extension MATH of MATH to ensure that MATH and MATH. In particular, choose MATH large enough that MATH and take MATH. We may also assume that MATH contains MATH, and that MATH has a MATH-semiunit basis, by enlarging MATH suitably. We construct a convergent sequence MATH over MATH such that MATH for each MATH, as follows. First set MATH. Given MATH, if MATH has no coefficients of norm at least REF, the proof is complete and we stop. Otherwise, write MATH. If the smallest positive index MATH for which MATH gives MATH, let MATH be the index, let MATH be a matrix over MATH whose entries are MATH-semiunits such that MATH, and put MATH. In this case we say MATH is of the ``first type". Otherwise, let MATH be the smallest positive index at which MATH is realized. By construction, MATH. Thus the lowest order terms of MATH form an invertible matrix over the ring of elements of MATH of MATH-norm REF modulo those of norm less than REF. That ring is isomorphic to MATH, and so is a principal ideal domain. Thus by REF , the matrix can be factored into elementary matrices. By lifting each of these into an invertible matrix over MATH (possible because MATH admits a MATH-semiunit basis) and multiplying together, we get a matrix which we again call MATH. In this case we say MATH is of the ``second type". In both cases, we set MATH. Note that MATH by virtue of the fact that MATH. As a consequence, MATH, since if the leading term on the right side is contained in MATH, that on the left is the corresponding entry of MATH, which has smaller MATH-norm by the choice of MATH and the bound MATH. In particular, we have MATH, so the argument can be continued. We now show that the MATH converge in the NAME topology on MATH. Before doing so, we construct an upper bound on the size of the coefficients of MATH of positive index that will be stable under conjugation by MATH. Namely, put MATH. For MATH in the value group of MATH, let MATH. Let MATH be the largest value such that MATH, and extend MATH to a smooth increasing function MATH on MATH which is identically REF for MATH. Now define MATH . Then it is easily verified that MATH for all MATH, MATH as MATH, and MATH for MATH. (The second condition follows from the inequality MATH obtained by bounding MATH by MATH.) By construction, MATH for all MATH. Moreover, because MATH for all MATH, it is easily verified by induction on MATH that for MATH, we have MATH for all MATH and MATH. Now observe that MATH takes any given value only finitely many times. To wit, if there were a smallest value that occurred infinitely often, the corresponding values of MATH would form a decreasing sequence (since the MATH-minimal terms of MATH would not change between these values of MATH, so MATH would decrease on these values). But MATH lies in the (discrete) value group of MATH, contradiction. Thus MATH takes any given value finitely many times, and so goes to infinity with MATH. Since MATH, MATH also goes to infinity. On the other hand, since MATH and MATH as MATH, MATH as MATH. By the same argument as in the previous paragraph, MATH for MATH. Therefore the MATH form a NAME sequence in the NAME topology, and their limit MATH exists. We finally prove that MATH has no coefficients of norm greater than or equal to REF. We first prove there are no coefficients of norm greater than REF. Put MATH and MATH. If MATH for some MATH, let MATH be the smallest index maximizing MATH. Then for some MATH, the coefficient of MATH in MATH has norm MATH for all MATH. In fact, we can choose MATH large enough so that MATH is maximized at MATH for all MATH, by finding MATH such that MATH for all MATH, and choosing MATH so that MATH for MATH. For MATH sufficiently large, MATH cannot be of the first type, since MATH for all MATH of the first type. So there exists MATH of the second type. By construction, MATH and MATH, contradiction. We conclude that MATH for all MATH. The proof that MATH for MATH is analogous but simpler: assuming the contrary, let MATH be the smallest index such that MATH. Then for MATH sufficiently large, MATH for MATH; but then any such MATH would be of the first type with MATH, and we would have MATH, contradiction. In conclusion, we have that MATH has entries in MATH and MATH, as desired. |
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